Understanding the Binomial Theorem

Now we've got to the point where we can talk about the last. Topic of the course, the topic that I've entitled at the top of the screen, the binomial theorem. Now I'm going to do this in three parts so we can sort of build ourselves up to the theorem itself. The first part will be to talk about the binomial coefficient symbol, because this is a new symbol you haven't seen. seen before. When I'm finished with this, we'll talk about something called Pascal's triangle, which you may have heard of, and then we'll go on to the binomial theorem. But let's do this first, the binomial coefficient symbol. And let's go ahead and define it, and then we'll practice a little bit. This won't take long, and that'll give you a little bit of preparation for what's coming up. Definition. If, and we're talking about integers now, if n and j are integers, With J being limited, it is between 0 and N. Then we're going to define this new funny looking symbol. Then we define. And here's what it looks like. N and then J below it and big parentheses around it. This is often pronounced N choose J or we might say N taken J at a time. Either one of those phrases are used. And what is this defined to be? After all, this is just a symbol. What does it actually mean? It is defined to be n factorial over j factorial times n minus j, quantity factorial. So you take the top number here. here and write its factorial, the bottom number and its factorial, and then n minus j is the difference between the two, and it's always the upper minus the lower, so it's n minus j factorial. This is the definition of this symbol called the binomial coefficient symbol. First of all, let me tell you that I will have to wait till I get to the binomial theorem to explain why it's called this binomial coefficient symbol, and let me make another note about this. We're going to limit ourselves. Even though it is defined for all integers, we'll limit ourselves to positive integers n. So that's all we'll look at in this course. Although it is interesting to note that not only will the binomial theorem work for positive integers n, it will also work for negative and for non-integers, which is a famous result in history. But for us, we'll go ahead and just... stay with this. And as I said, you might call this N choose J. That perhaps is the shortest phrase I know for talking about that symbol. This symbol also you may see somewhere as written N C. Now I'll tell you that the C stands for combinations, but don't worry about that. We will not be talking about probability in this course. But you see the N and the J appear. So that allows people to write this all in a single line. But we will not be using this. So let me just cross that out, all right? We will use this symbol as it is there. However, on your calculator, that symbol won't appear either. And if you're using a calculator, let me go ahead and give you a few notes here that are worth mentioning. One is, there may be a key that looks like NCR on your calculator. More likely, it's a key or a menu item. on your calculator that has the same name. If you want to go ahead and compute the binomial coefficient using your calculator. So I just wanted to warn you about that. The second note is a misconception students often have the first time they see this symbol. That's n over j with big parentheses, but there's no division symbol in between. So this is not n divided by j. It is not division. So be careful there. That's a tendency you have to write things out with division. Don't do that. This is a new symbol with no division. And for the purposes of our calculations below, let me remind you of something when we talked about factorials. Remember. that n minus j quantity factorial is not equal to n factorial minus j factorial. So this difference does not spread itself out. All right, well, let's just do a couple of examples to show you how one might calculate this. How about 3 choose 1? Well, by definition, that's 3 factorial over 1 factorial. times 3 minus 1 factorial. Well let's see what we have here. That's this bottom part I'll write first. The 1 factorial recall is just 1 so I won't rewrite it. 3 minus 1 is 2 factorial. The top you may recall you can factor out the highest number 3 and what's left will be 2 factorial. factorial. That was a famous result we had when we talked about factorials. Well, with two factorial over two factorial, that's one. We're left with three here. So three choose one is equal to three. And let's do one that's a little bit bigger. to get a little more dramatic here. How about 65 choose 15? There's a big one. What would that be? By definition this is 65 factorial. You really wouldn't want to write that one out very far. Over 15 factorial and then what goes here is the difference between the two. That is to say 65 minus 15 and I will write it directly as 50 factorial. Now, How can I simplify this? Well, the trick here always is to look at the factorial on the top and look at the biggest one on the bottom. The biggest one is 50. So I will write the top as 65 times 64 all the way down to... 51 times 50 factorial. Now that also follows from the same rule that gave us this. You can factor out as many terms as you want and the rest is factorial. On the bottom is 15 factorial, 50 factorial. Well with 50 factorial over 50 factorial that becomes 1 and the answer is then 65 times 64 times dot dot dot down to 51 over 15 factorial and whether you go any further is up to you. The number you get is rather large on the order of 10 to the 14th, but the point I wanted to make here is that this is the definition, this is how you might simplify it, and we'll see more of this, and you can stop wherever it's convenient for you to stop, because nobody wants to multiply this out by hand, certainly. Well, there's a fact about binomial coefficients. Remember, fact is just our name for a little theorem. That's going to be quite helpful later. So let me write this out. For positive integers, little n, and of course you know what that means. That's meaning... n is in the natural numbers. Those are the positive integers after all. And are j usually restricted? j greater than or equal to 0 less than or equal to n. In that case we have the following statement. n choose j is is the same thing as n choose n minus j. Aha! Now that will turn out to be extremely handy. And another way to say this with our traditional boxes here that we've introduced in this course, I could say n choose box is equal to n choose n minus box. And that will also be a way to remember this later. Where the box can be filled by anything, J here, but it can be filled by any variable or expression. Okay, well this is so easy let's go ahead and do the proof. I don't like giving you formulas without a reason. All that is required here is that we write down the definitions of these two and you'll see that they're the same. So let me start on the left. n choose j by definition is n factorial j factorial n minus j quantity factorial. But now I'm just going to rewrite the bottom. The top will remain n factorial. On the bottom, I will reverse, first of all. I'll write n minus j factorial out front. That doesn't change anything. And now look how I'm going to write j. Instead of writing j as j, I'm going to write it as n minus n minus j. Now, of course, n minus n minus j is n minus n. That's zero. And n minus minus is plus j. So this is j. All right, so I now have n minus n minus j factorial. Now, why on earth would anyone want to do that? Only if they're trying to make a point here. The point is that now this is exactly n over n minus j. Because remember how this definition goes. You put n factorial on the top, n minus j factorial on the bottom, and this is the difference between those two. So this handy fact has now been established, and we are going to use this later. It will allow us to make some very, very important simplifications. Now, there's a corollary I'll write down just for the record. In particular, let me show you two special cases. They occur so often it's worth mentioning them. If you take n choose 0 and the n is anything, you will always get 1. And that will always be the same as n choose n. Because, what do we have here? This is j, and this is n minus j, n minus 0 is just n, of course. Those will both be equal to 1. Likewise, n choose 1 will always be n, and that will be the same as n choose n minus 1. Now that comes right out of that theorem, that these two are equal. The fact that they're equal to 1 and n... I will leave to you to check. And I'll just show you an example to show you how that might work. If I took 5, choose 0, for example, that's 5 factorial, 0 factorial, and 5 minus 0 factorial. 0 factorial was conveniently defined to be 1, remember. So this is 5 factorial, and 5 minus 0, of course, is 5. So 5 factorial over 5 factorial is 1. And so you can go ahead and check this out if you like. The important thing I want you to remember from this section, apart from the definition of the binomial coefficient, is this fact here that n choose j is equal to n over n minus j. Very, very important to remember that. Well, on that note, let's stop and move on to the next piece of information I want to give you before the binomial theorem. This will be about Pascal's triangle. Now it's my great pleasure to talk about a very famous object from history called Pascal's Triangle. And as you'll see, the name is given to it because Pascal did a great deal of work on it and derived a great deal of theorems from it. But he was not the first one to ever see it and discuss it in mathematics. So let's go ahead and talk about what this might be. First of all, I want you to consider the following. Infinite triangle. Of course I won't be able to draw all of it, but I can go ahead and give you the indication of what it might look like to begin with. And then we'll talk about what we see here. Here's the triangle. One. 1, 1, written that way, and then 1, 2, 1, and then 1, 3, 3, 1. And there's actually a rule here for computing these. You may or may not know it. I'm just going to continue here until I run out of space. And it looks like I have just done so, so I'll just put dot, dot, dot there. There is a triangle of numbers. This is a triangle traditionally referred to as Pascal's triangle. You will find out that there are other people that have certain... claim on this. I would like to label a few things here. This first line is called row 0, the one with the single 1 in it. The next one is row 1. So we're starting with 0 and numbering it sequentially. So the next will be row 2, and the next one will be row 3, the next one will be row 4, and that will continue. So we have these rows of numbers being presented to us here. There's another thing to observe from this, that this is vertically symmetric, If I start drawing a vertical line down the center here, you'll see that on either side the numbers are identical. Now these are all important things to observe and let's write them down so we can talk about them. Observe the following. That we have, as I said, a vertical axis of symmetry. We have vertical axis of symmetry. And vertical axis of symmetry. That's that line down the center. Also, you may have noticed that each row begins and ends with the number 1. Okay, let me bring back the triangle here so you'll see. There's the triangle. Every one of these rows begins and ends with the number 1. Also, you may have noticed, and we'll go back and look at this in a moment, and this is the most important observation. This is how you draw the triangle. This is how you generate it. Each row is generated from the previous row, the one above it, the previous row. By the following. If you have a row going along, and here's a term in the row, and here's another term in there, and it continues onward. And here is the row below. And we get to the term that's in the middle there. How do we get from there to here? Well, the idea is you add these two and you get this one. So what you've got is these two adding to give you the one directly below it. Now let's go back to the triangle so you can observe that yourself. For example, if I'm looking for this 3, you see that 1 plus 2 equals 3. If I'm looking for the 6, 3 plus 3 equals 6. If I'm looking for the 10 here, 4 plus 6 equals 10. That works throughout the triangle. That's how I was able to generate it as I was writing it. So, this is the most important part of this triangle. But, this triangle, as I said, is one that's been around for a very long time. And before we go on and talk about its properties, I want to give you a little bit of its history. So, I'm going to show you some pictures now. Here's the first one. This is a picture of Pascal's triangle due to Jordanus de Namur, around 1225 AD. This appeared in one of his writings, this triangle as it stands here. And you can see that this is Pascal's triangle, isn't it? There's the ones running down the sides. This is 2. 1 plus 1 is 2. This 3 is 1 plus 2 is 3, etc. Let me show you another picture that appeared in his work also, this one. This is even more dramatically Pascal's triangle. There's a 1, 1, 1. There's the 2. There's the 3s. There's the 4. There's the 6 in the middle. 3 plus 3 is 6. So this also appeared about 1225. Giordano de'Namore was a European mathematician of the time. Now let's go across the world to China. In 1261, Yang Hui. had this in a book that he wrote. And this is exactly Pascal's triangle. Look at that. You can also learn Chinese numerals as you watch this. These horizontal lines evidently represent one. There's two. There's three. This is must be. be what six is, that's what four is, etc. So you can see this directly from his book in 1261. So this was known across the world. This was known in China. Let me show you another famous picture from China that's off. often presented in textbooks you'll see. This is from a mathematician named Chu circa 1303 and this is nicely drawn and you can see that this goes much further than the other one. And I know this is a little hard to see because these are a bit small on the screen so I've expanded and blown up this upper part. So let's go ahead and look at that. Here's the upper part detail. Still 1303 and Chu but now I've blown it up. And again you see the ones here, there's two, there are the three. trees. This is how six was drawn in the Chinese numerals of this date. And here's four, four lines and four lines here and so on. So Pascal's triangle here, and you can also see that the lines are drawn in for how the next term is created from the previous two. So that's quite nice. Pascal's triangle was known across the world. And so this is not an isolated event that Pascal happened to look at. This was known a very, very long time before him. Pascal Pascal lived in the 1600s, so this has been around for quite a while. And besides, this is pretty, so I wanted to show it to you. Now we've looked at Pascal's triangles in one sense. Now let me show you something else. Now consider another triangle. Another infinite triangle. Infinite triangle. Okay, here's the one I want you to look at now. And we'll use that new binomial coefficient notation we looked at. Here's the triangle. Put the 0 out of 0 binomial coefficient at the top. And then the 0 out of 1, and then 1 out of 1 here. And then 0 out of 2. Or 2 choose 0 if you like, or 2 choose 1 here, and 2 choose 2 here. And then let's do the 3s, if we can get them in. 3 choose 0, 3 choose 1, 3 choose 2, 3, 3 here. And I probably can get in the fourth one. 4, 0, 4, 1, 4, 2, 4, 3, and 4, 4. And then this continues also out to infinity. Observe that in these the row number is the top number of the binomial coefficient. You see there's 0, 1 row 1, 2 is row 2, 3 is row 3, 4 is row 4, so that's easy to check. And we have this vertical symmetry going on again, but now we can prove that that's true because recall we had a fact that told us exactly Exactly that, that n choose j is equal to n choose n minus j. That says that this one over here equals this one over here. This one and this one are equal. The two on either side here are equal. So we get the vertical symmetry due to this result here. Otherwise, it might not be apparent to you that you have vertical symmetry. So let me write down a couple of things, and then we'll ask the question about how that triangle is generated. So observe, by the fact, which I quoted on the side of that page, we did in the last section, there is a vertical axis of symmetry. Vertical axis of symmetry, as Pascal's triangle does have. Notice that each row begins with a vertical axis of symmetry. And ends with 1. Now how can you see that from the binomial coefficient? Remember, 1 is n0. It's also nn. And in that triangle that I just had up here, notice that the ones along this side all have zeros. on the bottom and the ones on this side always have, all have the top and bottom the same. So these are all ones and these are all ones. And by the fact once again we have the vertical symmetry across the axis. Well there's only one more thing to check and if it does turn out to be true then that triangle and Pascal's triangle are the same. So the question is does the row generation. Does the row generation rule work too? Well, the answer is yes. And now I'm going to prove it to you. And once I've done that, you'll see that Pascal's triangle and the triangle with the binomial coefficients we've been looking at are, in fact, the same triangle. Theorem. Here's the statement that says they're the same. The two triangles, triangles of numbers that is, triangles are the same. They are the same since... We've already seen the symmetry is the same. We've seen they start and end the rows the same way. In fact, the row generation technique is the same since the row generation rules are identical. Now what do I mean by that? Meaning, let me go ahead and show you pictorially here. Meaning that on the one hand we had dot, dot, dot, and we had a box in Pascal's triangle. And if you added it to the next element, then you got the one directly below. Right? We had this going on. Okay? That was the Pascal's triangle rule that I stated by showing you earlier. Then the rule we have now is going to be because n, j minus 1. 1 plus NJ. See, that's the next term as you go from left to right. The N say I'm in the Nth row. The J minus 1 to J tells me that these are terms that are right next to each other. That equals the N plus 1 row and the Jth term in it. So that is the binomial coefficient version of this rule for row generation. So this is row N that I'm talking about here. And this one is rho n plus 1 here and here. And this was Pascal's triangle, and this was the binomial coefficient triangle. And by showing these two rules are the same, which I'm about to prove by proving this statement over here, this starred statement, by proving that, I will show, therefore, the triangles are identical. And so I can interpret them either way, which will make a big difference for us when we're talking about the binomial theorem, which is coming up. Now here's the proof. And let me be frank about this. This is an exercise in factorials. Exercise in factorials. So, you may have to listen to this more than once to catch all the factorial cancellations. I'll try and show them all to you neatly, but this is a bit complicated with the factorials. n choose j minus 1 plus n choose j. Okay, this is... The left-hand side, here's the way I'm going to do this now. I'm going to take this left-hand side and show it's equal to the right-hand side here underneath my head, the 2n plus 1j. I'm going to show this is equal to this. So I'm going to start out with the left and hope I end up with the right. That's what I start out with. Well, the first thing I can do is write out the definitions of both of these using factorials. So the first one is n factorial over j minus 1 factorial. And then the difference between these two. that's n minus j minus 1. So I'm going to write that as n minus j and then plus 1 running the minus through. So n minus j plus 1. And that's factorial of course. Plus n factorial for the next one and then j factorial and n minus j factorial. So all I've done is write down the definitions. Now what I want to do is combine these into a single fraction. In order to do that I've got to find the common denominator. So what I'm going to do is look at what I have underneath here and see if I can discern what that ought to be. Now on this side over here, n minus j plus 1 is one number past n minus j. It's n minus j plus 1 after all. So this part here can be rewritten as n minus j plus 1 times n minus j factorial. I'm just taking the first number of the factorial out. Likewise here, the j can be written as j times j minus 1 factorial. Now why did I bother doing that? I did that because I had a j minus 1 factorial over here, and over here I had an n minus j factorial. And now with these observations, move my equal sign down a bit here, I can go ahead and write these out so that they look a little bit more the same. This one I will write out with these parts. So here's n factorial still on top. j minus 1 factorial and then n minus j plus 1 the number times n minus j factorial. KELOLAND dot com. plus, writing this out, n factorial, and then I have j here, j minus 1 factorial, and n minus j factorial. Well, that's good. Now, these denominators are still not the same. What's missing? Well, let's see. They both have j minus 1 factorial. They both have n minus j factorial. This one has an n minus j plus 1 that this one does not. So, let's go ahead and give it to it. So, multiply top and bottom by n minus j plus 1. Now this one on the right has a J that the first one does not. So we will multiply the top and bottom here by J. That will give us a common denominator here and that will take us to the next page and we can write these out as a single fraction now. So our common denominator will be the j, the j minus 1 factorial, the n minus j plus 1, and the n minus j factorial. And on the top we have j times n factorial. factorial plus that n minus j plus 1 times n factorial. And that's what happened on top. Remember what we had previously here. So the j times n factorial and here it's the n minus j plus 1 times n factorial. So finally we're at this stage and now the simplification is much easier. I can recombine these two into a single factorial as I can do here. In the top I can simplify by seeing that n factorial is common and factor it out. So n factorial times j plus n minus j plus 1. And the bottom becomes simply j factorial times n minus j plus 1 factorial. Well, the top is even more simple. simple. Notice j and minus j cancel. So I'm left with, if these two add up to zero, I'm left with n plus 1. Well, n plus 1 times n factorial is n plus 1 factorial over j factorial. And this one I'm going to rewrite n minus j plus 1. I'm going to reorganize it as n plus 1 minus j factorial. So I just reversed the positions of the minus j and the 1. But now you immediately read this by definition. This is n plus 1 j, the binomial coefficient. Because remember how we do this? n plus 1, n plus 1 factorial, j, j factorial. And this is the difference, n plus 1 minus j factorial, which it is. So to take us back through this, what were we trying to show? We were trying to show that this object here equals this object here. And we have now done that. And that will be the result that we want to use. That is the result from the theorem. Let me bring the theorem back so you remember it. That says that the Pascal triangle and the binomial coefficient triangle are the same. These are the same triangle because they have the same row generation rules. And it is in the form of binomial coefficients that we will see this triangle again. And that will be coming up in the next segment, which is finally the binomial theorem. a major result. And now we come to the major result which is the title of this page at the top of your screen, the Binomial Theorem. Finally, we get to this, the binomial theorem, how to expand a binomial, x plus a, to the nth power. This is a major theorem. It is a nice piece of work to end this unit. And we've got all the preparation now from having looked at the binomial coefficient symbol and Pascal's triangle. So let's put it all together. Okay, first of all, on the binomial theorem, how to expand x plus a. To the end, what I've done is I've written out the first few powers. So we can detect what sort of pattern might be here. So I've got the powers from 0 up to 4 here. And I've written them out in a suggestive form. It's a triangle, which should suggest something. So I've got 1. I've got x plus a, x squared, plus 2x, a plus a squared. And then the x cubed and the x plus a to the fourth row here. And then, of course, this continues forever. Now, I don't know if you see anything here yet, but there are some patterns that will become familiar. And I think what I'm going to do is keep this and come back to it because I want to organize this differently. It's a little hard to see what's going on. I like to organize it this way, and I'll put it down this way. Can we detect a pattern? Okay? And there may be several, but let's see if we can pull all of these together. And you know, when we're done with this, I'll make an observation that I think you'll find interesting. X plus A to the 0. Okay? Let me write this vertically the following way. Let me put the coefficient 1 here, and then put X to the 0, A to the 0. Okay, x plus a to the 0, of course, is 1. So I'm writing out 1 in a rather elaborate fashion. 1 and 1 and 1 in this fashion. Now, you'll see from the next two what I'm trying to do here and what the pattern is. Let's look at x plus a to the 1 power. And let me write that out as follows. The coefficient first, 1, x to the 1 times a to the 0, and below it I'll write plus the next one whose coefficient is 1, x to the 0, a to the 1. Now, I'm going to write the coefficient first, 1, x to the 1 times a to the 0, and you know that a plus, x plus a to the 1 is just x plus a after all. And again, this looks more elaborate than it needs to be. But as we get on, you'll see that all of this is necessary. This is x to the 1 again. That's a 1, that's a 1, so this is just x. There's a to the 1, so that's a, and these are just 1. So it is x plus a. Okay? So there's the first one, and here's the second one. Now it begins to look more interesting. x plus a squared. Let me write this out this way. Coefficient. 1, x squared, a to the 0, plus coefficient 2, x to the 1, a to the 1, plus coefficient 1 again, x to the 0, a squared. Now don't worry too much about the coefficients that I have boxed just yet, but I think you should observe a pattern for the x's and the a's. It's most evident in this last one, but it is true in the first two also. In the last one, notice that the power that I'm raising the binomial is 0. binomial 2 is 2, the first power of x is 2, and it goes down to 0, 2, 1, 0. Notice that simultaneously the a's go from 0 up to 2. So they're going in reverse order. And we'll get to the coefficients later. Same thing happens here. 1 to 0, 0 to 1. And of course here they're both the same because there's only one place to go. Now let me continue this pattern a little bit further so that you'll begin to see it. I'll do it for the cube and for the fourth power. Okay, x plus a cubed. This can be written as 1x cubed a to the 0 plus 3x squared a to the 1 plus 3x to the 1a squared plus 1x to the 0a cubed. Now, you may recall that this is exactly how it looks. Notice that the x powers go down from 3 to 0, the a powers go up from 0 to 3. Nice pattern. You still may not recognize the numbers yet, but don't worry, I boxed them in. for a reason so we can come back and look at them. Let me do one more just to set this in your mind. X plus A to the fourth power is equal to 1 times X to the fourth A to the zero. And apart from the coefficients, you should be able to guess all the coefficients. all the x's and a's now. x cubed a to the 1, that's a 4 there, and a 6. x squared a squared plus another 4, and this is x to the 1 a cubed plus 1, and this is x to the 0 a to the 4th. And again, you notice the x powers go down from 4 to 0, and they go up from 0 to 4 for the a's. And then there are these coefficients, and then this course continues on out to infinity, as many times as you want to take the binomial to a power. Well, let's write down what we've observed so far, and then go back and see if we can pick up what we can say about the coefficients. Observe. Well, Well, the first thing I observe is that each terms, I didn't mention this, but each terms powers of x and a add to n. Now, did you notice that? Let's come back here. We noticed that the powers of x go down and the powers of a go up, but did you notice that the sum is always 3? Here, 3, 3, 3, and 3. Same thing here, add to 4, 4, 4, 4, and 4. So that's an observation we could have made also. So each term's powers of x and a add to n. The x powers go from n down to 0. So that's down. The way I organized it, and the a powers go from 0 up to n. Okay, those are the observations about the powers of x and a. What about the coefficients? Well, the coefficients have the following. pattern and let's see if we go back and pick those up. First of all we have a 1, a 1, 1, a 1, 2, 1 and then continuing here we had a 1, 3, 3, 1, a 1, 4, 6, 4, 1. If I wrote those out just as they appeared, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, and then 1, 4, 6, 4, 1, and continued out to infinity. Do you happen to notice what that is? That is exactly what we called previously Pascal's triangle. Or if you like, the binomial coefficient triangle, and now you know why it's called the binomial coefficient symbol and the binomial coefficient triangle, because the numbers are binomial coefficients after all. This is exactly... Now this is not the way we drew it, but you can see that the rows here, which are now columns, are the same. There's 1, there's 1, 1, there's 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1. In fact, this is the way that Pascal actually drew the triangle. drew the triangle when he was working on it originally. So he must have been thinking something along the lines of the analysis we just did. OK, with all that preparation, we can finally write down the theorem that we've been trying to get to. Theorem. This is the famous binomial theorem. Okay, just like that and it goes like this if X and A are in the real numbers. So those are the two numbers that are going to be part of the binomial. And N, for us, will be in the natural numbers only. We'll talk about powers like 1, 2, 3, etc. Now, I did use the power 0 in my examples, and we could extend this to include 0, but we'll usually be looking at powers that are higher than 2 and 3 and 4 because those are the ones we really want to expand. And we don't know how to do those easily any other way. All right. Then, here's what we get, x plus a to the n. What could this possibly be if you try to multiply it all out? Here's what it is. It is the summation of the n choose j's times x to the n minus j times a to the j, where the j will go from 0 to n. Okay, so there'll be n term, there'll be a number of terms in this, actually n plus 1, and these are the way the individual terms will look. The coefficient here is from Pascal's triangle or the binomial coefficient triangle. This is the n, this is the power of x and notice that we start out with j equals zero. So the power of x starts out n and then it continually goes down as we saw. A to the j starts out with j equals zero and will go up as we saw. And so this is the result that we've been trying to get to. And let me go ahead and write this summation out a bit so you'll see what we're talking about. If I write out a few of these terms, I start out with n choose zero. x to the n, because in this case, a to the 0, of course, is 1. plus n choose 1, x to the n minus 1, a to the 1, plus dot, dot, dot, and then I'll give you what the last two terms look like, dot, dot, dot, plus n choose n minus 1, x to the 1, a to the n minus 1, plus the last one, which is n choose n, a to the n, and the x to the 0 is really there, but I'm not writing it because that's just equal to 1. Now I won't prove this. Okay, we've set the scene for this. We've analyzed quite a We've looked at the coefficients, the powers of x and a, and I think we've set up how this ought to go. We won't prove this, but we will talk about this in detail. So let's go ahead and first of all rewrite that with maybe the boxes notation. Maybe that's helpful to have. x plus a to the n would be, now here I'm going to use the box for j. So it'll be n choose box. Times x to the n minus box times a to the box. Where the box goes from 0 to n. So that's a nice way of seeing where that index number takes its positions. Put a box in there because that's unusual and it catches your eye. So there's a box here, a box here, a box here. That's where the index goes. What's a typical term look like? This is sometimes a question you're asked. What does an individual term happen to look like? Well, the box plus first. term. Remember we're starting with the box equals zero and we're going to n, so that means we have n plus one things added. So the n plus first term or the box plus first term will be this object up here. n choose box. x, times x to the n minus box, times a to the box. There it is. That is the n plus box plus first term for the box varying from 0 to 1 to 2 up to n. Okay? I've been saying n plus 1. I've been trying to avoid that. Say box plus 1 because that really does stand for the j. Okay, one more thing I want to recall before we do an example. Recall the fact that we had about the binomial coefficients that I said would be useful, that n choose j is the same thing as n choose n minus j. Now we're going to use this in our examples coming up, so I wanted to make sure that was fresh in your mind. So a typical term looks like this, and this is a fact we will be using. Example, let's go ahead and expand a binomial that we would prefer not to write by hand. X plus 2, say, to the fifth power. Now, yes, we can go ahead and expand that by hand, but that would be tedious. And as the powers get higher, it gets even worse. So, let's write down that X plus 2 to the fifth by the theorem is the summation I'll 5 choose J. Now notice the way I'm writing this. 5 is my exponent here, so that's what's going to be the top of my binomial symbol. The bottom is always J, so I do 5 choose J to remember that. And then I come back here and write J equals 0 to 5, and then I put in the rest of it. X to the 5 minus J times 2 to the J. And in this case, if you write it all the way out, Let me put a line here for clarity on the screen. 5 choose 0, and we have x to the 5th, because x's are going to start high and go down, and the a's start at a to the 0 and go up, plus 5 choose 1. x to the fourth times 2 plus 5 choose 2 x cubed times 2 squared and let's come down here 5 choose 3 x squared times 2 cubed you see the X's are going down and the twos are going up 5 4 X to the 1 times 2 to the fourth plus finally 5, 5 times just plain 2 to the 5th because here is where x is equal to the 0 power. Okay? Advice. Use that fact. That we had on the previous page. That I'll even write again that n choose j is equal to n choose n minus j. Because what this does is it allows you to use the symmetry of Pascal's triangle or the binomial coefficient triangle. Which means. that 5, 2, 0, and 5, 5 are going to be the same number. We happen to know that's one from our experience. But the point is they're the same. As we come in, these two are going to be the same. And then finally these two will be the same. And if there were an an odd number, there'd be one in the middle that we'd have to calculate. But at least we can use all of this symmetry to make our work go a little bit more quickly. So writing some of that out, here 5 choose 0 is the same as 5 choose 5, and we already learned earlier that that was 1. And 5 choose 1 is the same thing as 5 choose 5 minus 1, which is 4. That was equal to 5, and this is Recall because we did this earlier. I showed you examples of that. The only one I don't know is 5 choose 2, which is the same thing as 5 choose 3. So let's calculate it. That's 5 factorial. I'll do the first one. 2 factorial, 3 factorial. Once again, I'm going to take the bigger one, the 3 factorial, into the 5 factorial. That will leave me with 5 times 4 on the top over 2 factorial, because the 3 factorial over 3 factorial, the one that's inside the 5 factorial, will cancel with this one to be 1. That's 20 over 2 factorial. 2 factorial is just 2, so that's 10. So finally... I can write down what I wanted to write down. What is the expansion of x plus 2 to the 5th? Well, coming from the previous page, it's x to the 5th plus 5 times x to the 4th times 2 plus 10 times x cubed. times 2 squared plus 10 times x squared times 2 cubed plus 5 times x times 2 to the 4th plus 2 to the 5th. And we can go ahead and finish that off by multiplying all these numbers out. out the 5 times the 2, the 10 times the 2 squared, 10 times 2 cubed, 5 times 2 to the fourth and multiplying out 2 to the fifth to get finally x to the fifth plus 10x to the fourth plus 40x cubed. cubed, plus 80x squared, plus another 80x, plus finally 32. And that is the expansion of x plus 2 to the 5th that the binomial theorem allows us to do. In order to do this, we needed to calculate, let me bring back the previous sheet here, when we started this. We have to be able to calculate all the binomial coefficients that appear. Plus, we have to look at the powers, in this case the powers of 2 that appear, that might need to be calculated. It's time for you to try a problem. Let me pose this one to you. Looks a little different from the one I just did, but I gave you a hint so I don't think it'll be too bad. Expand 2y minus 3 to the fourth power. A little less power, so it's not quite as much work. And you have a binomial here, but the theorem talks about x plus a. So my hint is to read this as 2y plus a minus 3. So the 2y can act like the x would act, and the minus 3 will act like the a would act in the theorem. So you go ahead and try to expand this to the bitter end, and I'll come back in a moment and show you what I did. All right, here we begin. We're looking for 2y minus 3 to the 4th power. That's what we'd like to expand. Okay, well, simply applying the binomial theorem as it stands, this is going to be a summation of, and it'll be 4 choose j, and the j will go from 0 to 4 then, times the first object, which is 2y. to the 4 minus j power times the second object, which I realize is minus 3. So that's minus 3 to the j power. Okay, that's just writing down what the theorem tells me to write down. All right, I know I need to do a lot of work here, so let me write out the entire expression just so I have it in front of me. 4, 2, 0, and then I'll have 2y to the fourth. Of course, the other part is to the zero, so I don't need to write it. Plus... 4 choose 1, 2y quantity cubed times minus 3 to the 1 plus 4 choose 2. And now I'm going down in powers, 2y squared minus 3 squared plus 4 choose 3, 2y to the 1 power, minus 3 is going up now, cubed. Plus 4, choose 4. Minus 3 to the 4th. And, of course, here the 2y is to the 0 power. And let me run a line through here again so it will be a little easier for you to see. Okay? That is what that is totally multiplied out. Now you see what work I have to do. to figure all the binomial coefficients and then I have to do other multiplication. If 2y is cubed, I have to have 2 cubed times y cubed. And then all these various powers of minus 3 need to be dealt with. So, let's start with that. 4 choose 0 is the same thing as 4 choose 4 and that's 1, we knew that. And then 4 choose 1 is the same thing as 4 choose 3. We also knew that was 4. The middle one, the 4 choose 2, we don't know, so let's calculate it. 4 factorial over 2 factorial times 2 factorial, because this 2 is 4 minus 2. But notice that 2 factorial and 2 factorial are both equal to 2. And 2 times 2 is 4. And 4 will cancel with one of these up there, leaving me with plain 3 factorial. But 3 factorial we know is 6. So I now have that binomial coefficient finished. So, I would want to go further and multiply all of these minus 3's, pull the powers of 2 out. And so with that in mind, knowing that you've done this, I can now write out that 2y minus 3 to the fourth power is equal to 16, that's the 2 to the fourth, y to the fourth, plus 4 times 8 times minus 3y cubed, and I'll need to combine those. plus 6 times 4 times 9, I'll combine those, times y squared, plus 4 times 2 times minus 27 times y, plus 81 because that's 3 to the 4th, and minus 3 to the 4th is the same thing, of course. So finally I have 16y to the 4th minus 96y cubed plus 216y squared minus 216y plus 81. And there you have it, the expansion that you were looking for. And I hope that's what you got and that it wasn't too difficult. Further, I want to pose the following question, because you are very often asked this kind of question in problems. Which term... contains a particular power of k say x to the k more often someone will ask you a specific question like in the expansion of like the one we did 2y minus 3 to the fourth which term contains the power y cubed okay find what that term is find the coefficient etc Well, the answer is to recall what the typical term looks like and then translate it into what we need. The typical term of the binomial expansion, if you recall, was n choose box, remembering my boxes, times n minus box, x to the n minus box, times a to the box. Now what do we want here? We want the power of x to be k. So that means we want n minus box. to be our K. Well, then what is the box going to be? Well, it's clear that N minus K then is equal to the box. That's just a tad of algebra. So what do we have then? We know that the, what we're trying to do here is is get the term containing x to the k. So the term containing x to the k is now going to be equal to, let's see, putting n minus k in for the box, I have n, choose n minus k, and the n minus box is k, so this is times x to the k, which we were expecting, times a to the n minus k. And there it is. If you're looking for the term containing x to the k, there is that term. Now this can be handy if you have a specific question in front of you. For example, let me show you how this works. Here's an example. Find the coefficient, the numerical coefficient, find the coefficient of y to the 8th power in the expansion of 2y plus 3 to the 10th power. Now the idea here is that you do not want to have to expand this to find out that coefficient. You want to do as little work as possible. So, first note that the term containing y to the 8th, which is what we're looking for, is the same as that containing... Well, since we have the term 2y there, 2y to the eighth, of course. It's whatever contains that. So that's an observation. Then from above, the fact above that we just computed, from the fact above, with k equal 8 here and n equal 10, we have that n over n choose k. times x to the k times a to the n minus k, okay, is equal to 10 choose 2 times 2y to the 8th times 3 squared. And we know that we want the numerical coefficient of y to the 8th here, so we need to work some things out. So let's first of all deal with 10 choose 2. By definition, that's 10 factorial over 2 factorial, 8 factorial, because 8 is 10 minus 2. And then dividing the 8 factorial into the 10 factorial, I'm left over on the top with 10 times 9 over 2, because that's all the 2 factorial is. That's 90 over 2. That's just 45. We'll go to the next page and pick out the other numbers we need here. We need 2 to the 8th. 2 to the 8th is just 256, and finally, the 3 squared, of course, is 9. So, we end up with the final coefficient is going to be 45 times 256. times 9 all times y to the eighth. That comes out to be 103,680. That's the coefficient we were looking for right there. 103,680. That's the coefficient that is the coefficient of y to the eighth in the expansion of the binomial that was given. So sometimes questions like that are asked, and I wanted to show you at least one example. But I wanted to leave you with a puzzle, and I'll come back and show you how to attack this. You can attack this using the binomial theorem, which is kind of nice. And I like this. This is a pleasant puzzle. The question is, which is greater? 1,000 to the 1,000 power, or 1,001, a little bigger, to the 999 power, a little smaller? Which one of these is bigger? Now my hints are the binomial theorem helps, so you might consider using the binomial theorem. There's a new fact that I'm adding right now that I think will help you through this. It helped me in doing it. The fact is that n choose r is always less than n plus 1 to the i. our power. So just jot that down and use it if it seems appropriate. The final hint is that your calculator probably won't help. These numbers are just too big for the calculator to distinguish them. It may even say that they are the same. They are not the same. One is actually bigger than the other and I won't tell you which. So why don't you go ahead and try that and I'll come back in a moment with my solution. Well, has the suspense been building? Let's see what the answer is going to be. As I said, I'm going to use the binomial theorem here. So by the binomial theorem... It will allow me to break this up in a very handy way. I will take 1,001 to the 999 power and rewrite that as 1,000 plus 1 to the 999th power. Now, of course, it's a binomial, so I can use the binomial theorem. And to be specific, I'm talking about the case where n, the power, is 999, the x value is 1,000, and the a... the other part of the binomial is just one here. Okay, what would the binomial theorem tell us? Well, it tells us this is equal to a sum of these coefficients, 999 choose j, and then the j will go from 0 to 999, times, let's see, the first term is 1,000 to the 999 minus j, times the second term, which is just 1, to the j power. Now, of course, 1 to the j... It'll just be one, so we'll drop this out from the subsequent writings of this. But the very first thing I'm going to use is that fact that I mentioned on the previous page so that I can get this inequality. Let me remind you what the fact was. The fact was this, that n choose r is always less than n plus 1 to the r. Now, there may be other ways to do this problem, but this is certainly a very easy way that I saw. So I'm going to use that on the 999. choose J here and I'm going to say it's less than this object so this will make this less than the summation from J equals 0 to 999 of In place of this object here, this 999 choose j, I will put the other side of the inequality, the n plus 1 to the r. Now in this case, if n is 999, then this is 1000, and the r is j, so that's the jth power. So I have 1000 to the j times 1000 to the 999 minus j, and of course I won't write that 1 as I said I wouldn't. But now look what I have here. I have... something that can easily be combined into a single power. A power 1,000 to the j plus 999 minus j, which of course is just 999. So let me put that on this other page here. And just for the record, let me remember that my inequality is going that way, so I don't forget from page to page. And now that's going to be equal to the summation of, as I said, 1,000 to the 999, where j goes from 0. to 999. Now that is a constant. There is no J, the index letter, in there. How many terms are there? How many times do we add this constant in this context? Well, the J goes from 0 to 999. That's 1,000 terms. So let me write that down. 1,000 terms. And if that's 1,000 terms, then the inside is simply going to be multiplied by 1,000. So 1,000 times 1,000 to the 999. And of course, that immediately simplifies to 1,000 to the 1,000 power. So that was the other side of our inequality we were hoping to get to. So just to put it all together here, we've discovered that 1,001 to the 999, remembering the direction of my inequality, is less than 1,000 to the 1,000 power. So, I hope you found that enjoyable and an interesting little problem. That will be the end of Unit 7 for us, and from here you can go on to do a variety of other things.

The first part will be to talk about the binomial coefficient symbol, because this is a new symbol you haven't seen. seen before. When I'm finished with this, we'll talk about something called Pascal's triangle, which you may have heard of, and then we'll go on to the binomial theorem.

But let's do this first, the binomial coefficient symbol. And let's go ahead and define it, and then we'll practice a little bit. This won't take long, and that'll give you a little bit of preparation for what's coming up.

Definition. If, and we're talking about integers now, if n and j are integers, With J being limited, it is between 0 and N. Then we're going to define this new funny looking symbol. Then we define.

And here's what it looks like. N and then J below it and big parentheses around it. This is often pronounced N choose J or we might say N taken J at a time. Either one of those phrases are used. And what is this defined to be?

After all, this is just a symbol. What does it actually mean? It is defined to be n factorial over j factorial times n minus j, quantity factorial.

So you take the top number here. here and write its factorial, the bottom number and its factorial, and then n minus j is the difference between the two, and it's always the upper minus the lower, so it's n minus j factorial. This is the definition of this symbol called the binomial coefficient symbol. First of all, let me tell you that I will have to wait till I get to the binomial theorem to explain why it's called this binomial coefficient symbol, and let me make another note about this. We're going to limit ourselves.

Even though it is defined for all integers, we'll limit ourselves to positive integers n. So that's all we'll look at in this course. Although it is interesting to note that not only will the binomial theorem work for positive integers n, it will also work for negative and for non-integers, which is a famous result in history.

But for us, we'll go ahead and just... stay with this. And as I said, you might call this N choose J. That perhaps is the shortest phrase I know for talking about that symbol. This symbol also you may see somewhere as written N C. Now I'll tell you that the C stands for combinations, but don't worry about that.

We will not be talking about probability in this course. But you see the N and the J appear. So that allows people to write this all in a single line.

But we will not be using this. So let me just cross that out, all right? We will use this symbol as it is there. However, on your calculator, that symbol won't appear either. And if you're using a calculator, let me go ahead and give you a few notes here that are worth mentioning.

One is, there may be a key that looks like NCR on your calculator. More likely, it's a key or a menu item. on your calculator that has the same name. If you want to go ahead and compute the binomial coefficient using your calculator. So I just wanted to warn you about that.

The second note is a misconception students often have the first time they see this symbol. That's n over j with big parentheses, but there's no division symbol in between. So this is not n divided by j. It is not division. So be careful there.

That's a tendency you have to write things out with division. Don't do that. This is a new symbol with no division. And for the purposes of our calculations below, let me remind you of something when we talked about factorials. Remember.

that n minus j quantity factorial is not equal to n factorial minus j factorial. So this difference does not spread itself out. All right, well, let's just do a couple of examples to show you how one might calculate this. How about 3 choose 1?

Well, by definition, that's 3 factorial over 1 factorial. times 3 minus 1 factorial. Well let's see what we have here. That's this bottom part I'll write first.

The 1 factorial recall is just 1 so I won't rewrite it. 3 minus 1 is 2 factorial. The top you may recall you can factor out the highest number 3 and what's left will be 2 factorial. factorial.

That was a famous result we had when we talked about factorials. Well, with two factorial over two factorial, that's one. We're left with three here.

So three choose one is equal to three. And let's do one that's a little bit bigger. to get a little more dramatic here.

How about 65 choose 15? There's a big one. What would that be? By definition this is 65 factorial.

You really wouldn't want to write that one out very far. Over 15 factorial and then what goes here is the difference between the two. That is to say 65 minus 15 and I will write it directly as 50 factorial. Now, How can I simplify this?

Well, the trick here always is to look at the factorial on the top and look at the biggest one on the bottom. The biggest one is 50. So I will write the top as 65 times 64 all the way down to... 51 times 50 factorial. Now that also follows from the same rule that gave us this. You can factor out as many terms as you want and the rest is factorial.

On the bottom is 15 factorial, 50 factorial. Well with 50 factorial over 50 factorial that becomes 1 and the answer is then 65 times 64 times dot dot dot down to 51 over 15 factorial and whether you go any further is up to you. The number you get is rather large on the order of 10 to the 14th, but the point I wanted to make here is that this is the definition, this is how you might simplify it, and we'll see more of this, and you can stop wherever it's convenient for you to stop, because nobody wants to multiply this out by hand, certainly.

Well, there's a fact about binomial coefficients. Remember, fact is just our name for a little theorem. That's going to be quite helpful later.

So let me write this out. For positive integers, little n, and of course you know what that means. That's meaning... n is in the natural numbers.

Those are the positive integers after all. And are j usually restricted? j greater than or equal to 0 less than or equal to n. In that case we have the following statement.

n choose j is is the same thing as n choose n minus j. Aha! Now that will turn out to be extremely handy. And another way to say this with our traditional boxes here that we've introduced in this course, I could say n choose box is equal to n choose n minus box. And that will also be a way to remember this later.

Where the box can be filled by anything, J here, but it can be filled by any variable or expression. Okay, well this is so easy let's go ahead and do the proof. I don't like giving you formulas without a reason.

All that is required here is that we write down the definitions of these two and you'll see that they're the same. So let me start on the left. n choose j by definition is n factorial j factorial n minus j quantity factorial. But now I'm just going to rewrite the bottom. The top will remain n factorial.

On the bottom, I will reverse, first of all. I'll write n minus j factorial out front. That doesn't change anything. And now look how I'm going to write j.

Instead of writing j as j, I'm going to write it as n minus n minus j. Now, of course, n minus n minus j is n minus n. That's zero. And n minus minus is plus j. So this is j.

All right, so I now have n minus n minus j factorial. Now, why on earth would anyone want to do that? Only if they're trying to make a point here. The point is that now this is exactly n over n minus j.

Because remember how this definition goes. You put n factorial on the top, n minus j factorial on the bottom, and this is the difference between those two. So this handy fact has now been established, and we are going to use this later. It will allow us to make some very, very important simplifications.

Now, there's a corollary I'll write down just for the record. In particular, let me show you two special cases. They occur so often it's worth mentioning them. If you take n choose 0 and the n is anything, you will always get 1. And that will always be the same as n choose n. Because, what do we have here?

This is j, and this is n minus j, n minus 0 is just n, of course. Those will both be equal to 1. Likewise, n choose 1 will always be n, and that will be the same as n choose n minus 1. Now that comes right out of that theorem, that these two are equal. The fact that they're equal to 1 and n...

I will leave to you to check. And I'll just show you an example to show you how that might work. If I took 5, choose 0, for example, that's 5 factorial, 0 factorial, and 5 minus 0 factorial. 0 factorial was conveniently defined to be 1, remember.

So this is 5 factorial, and 5 minus 0, of course, is 5. So 5 factorial over 5 factorial is 1. And so you can go ahead and check this out if you like. The important thing I want you to remember from this section, apart from the definition of the binomial coefficient, is this fact here that n choose j is equal to n over n minus j. Very, very important to remember that. Well, on that note, let's stop and move on to the next piece of information I want to give you before the binomial theorem. This will be about Pascal's triangle.

Now it's my great pleasure to talk about a very famous object from history called Pascal's Triangle. And as you'll see, the name is given to it because Pascal did a great deal of work on it and derived a great deal of theorems from it. But he was not the first one to ever see it and discuss it in mathematics. So let's go ahead and talk about what this might be. First of all, I want you to consider the following.

Infinite triangle. Of course I won't be able to draw all of it, but I can go ahead and give you the indication of what it might look like to begin with. And then we'll talk about what we see here. Here's the triangle.

One. 1, 1, written that way, and then 1, 2, 1, and then 1, 3, 3, 1. And there's actually a rule here for computing these. You may or may not know it. I'm just going to continue here until I run out of space.

And it looks like I have just done so, so I'll just put dot, dot, dot there. There is a triangle of numbers. This is a triangle traditionally referred to as Pascal's triangle. You will find out that there are other people that have certain...

claim on this. I would like to label a few things here. This first line is called row 0, the one with the single 1 in it. The next one is row 1. So we're starting with 0 and numbering it sequentially. So the next will be row 2, and the next one will be row 3, the next one will be row 4, and that will continue.

So we have these rows of numbers being presented to us here. There's another thing to observe from this, that this is vertically symmetric, If I start drawing a vertical line down the center here, you'll see that on either side the numbers are identical. Now these are all important things to observe and let's write them down so we can talk about them.

Observe the following. That we have, as I said, a vertical axis of symmetry. We have vertical axis of symmetry. And vertical axis of symmetry. That's that line down the center.

Also, you may have noticed that each row begins and ends with the number 1. Okay, let me bring back the triangle here so you'll see. There's the triangle. Every one of these rows begins and ends with the number 1. Also, you may have noticed, and we'll go back and look at this in a moment, and this is the most important observation. This is how you draw the triangle. This is how you generate it.

Each row is generated from the previous row, the one above it, the previous row. By the following. If you have a row going along, and here's a term in the row, and here's another term in there, and it continues onward.

And here is the row below. And we get to the term that's in the middle there. How do we get from there to here?

Well, the idea is you add these two and you get this one. So what you've got is these two adding to give you the one directly below it. Now let's go back to the triangle so you can observe that yourself. For example, if I'm looking for this 3, you see that 1 plus 2 equals 3. If I'm looking for the 6, 3 plus 3 equals 6. If I'm looking for the 10 here, 4 plus 6 equals 10. That works throughout the triangle.

That's how I was able to generate it as I was writing it. So, this is the most important part of this triangle. But, this triangle, as I said, is one that's been around for a very long time.

And before we go on and talk about its properties, I want to give you a little bit of its history. So, I'm going to show you some pictures now. Here's the first one. This is a picture of Pascal's triangle due to Jordanus de Namur, around 1225 AD.

This appeared in one of his writings, this triangle as it stands here. And you can see that this is Pascal's triangle, isn't it? There's the ones running down the sides. This is 2. 1 plus 1 is 2. This 3 is 1 plus 2 is 3, etc. Let me show you another picture that appeared in his work also, this one.

This is even more dramatically Pascal's triangle. There's a 1, 1, 1. There's the 2. There's the 3s. There's the 4. There's the 6 in the middle. 3 plus 3 is 6. So this also appeared about 1225. Giordano de'Namore was a European mathematician of the time. Now let's go across the world to China.

In 1261, Yang Hui. had this in a book that he wrote. And this is exactly Pascal's triangle.

Look at that. You can also learn Chinese numerals as you watch this. These horizontal lines evidently represent one. There's two. There's three.

This is must be. be what six is, that's what four is, etc. So you can see this directly from his book in 1261. So this was known across the world.

This was known in China. Let me show you another famous picture from China that's off. often presented in textbooks you'll see.

This is from a mathematician named Chu circa 1303 and this is nicely drawn and you can see that this goes much further than the other one. And I know this is a little hard to see because these are a bit small on the screen so I've expanded and blown up this upper part. So let's go ahead and look at that.

Here's the upper part detail. Still 1303 and Chu but now I've blown it up. And again you see the ones here, there's two, there are the three.

trees. This is how six was drawn in the Chinese numerals of this date. And here's four, four lines and four lines here and so on.

So Pascal's triangle here, and you can also see that the lines are drawn in for how the next term is created from the previous two. So that's quite nice. Pascal's triangle was known across the world.

And so this is not an isolated event that Pascal happened to look at. This was known a very, very long time before him. Pascal Pascal lived in the 1600s, so this has been around for quite a while. And besides, this is pretty, so I wanted to show it to you.

Now we've looked at Pascal's triangles in one sense. Now let me show you something else. Now consider another triangle.

Another infinite triangle. Infinite triangle. Okay, here's the one I want you to look at now. And we'll use that new binomial coefficient notation we looked at. Here's the triangle.

Put the 0 out of 0 binomial coefficient at the top. And then the 0 out of 1, and then 1 out of 1 here. And then 0 out of 2. Or 2 choose 0 if you like, or 2 choose 1 here, and 2 choose 2 here. And then let's do the 3s, if we can get them in. 3 choose 0, 3 choose 1, 3 choose 2, 3, 3 here.

And I probably can get in the fourth one. 4, 0, 4, 1, 4, 2, 4, 3, and 4, 4. And then this continues also out to infinity. Observe that in these the row number is the top number of the binomial coefficient. You see there's 0, 1 row 1, 2 is row 2, 3 is row 3, 4 is row 4, so that's easy to check.

And we have this vertical symmetry going on again, but now we can prove that that's true because recall we had a fact that told us exactly Exactly that, that n choose j is equal to n choose n minus j. That says that this one over here equals this one over here. This one and this one are equal. The two on either side here are equal.

So we get the vertical symmetry due to this result here. Otherwise, it might not be apparent to you that you have vertical symmetry. So let me write down a couple of things, and then we'll ask the question about how that triangle is generated.

So observe, by the fact, which I quoted on the side of that page, we did in the last section, there is a vertical axis of symmetry. Vertical axis of symmetry, as Pascal's triangle does have. Notice that each row begins with a vertical axis of symmetry.

And ends with 1. Now how can you see that from the binomial coefficient? Remember, 1 is n0. It's also nn. And in that triangle that I just had up here, notice that the ones along this side all have zeros. on the bottom and the ones on this side always have, all have the top and bottom the same.

So these are all ones and these are all ones. And by the fact once again we have the vertical symmetry across the axis. Well there's only one more thing to check and if it does turn out to be true then that triangle and Pascal's triangle are the same. So the question is does the row generation.

Does the row generation rule work too? Well, the answer is yes. And now I'm going to prove it to you.

And once I've done that, you'll see that Pascal's triangle and the triangle with the binomial coefficients we've been looking at are, in fact, the same triangle. Theorem. Here's the statement that says they're the same. The two triangles, triangles of numbers that is, triangles are the same.

They are the same since... We've already seen the symmetry is the same. We've seen they start and end the rows the same way.

In fact, the row generation technique is the same since the row generation rules are identical. Now what do I mean by that? Meaning, let me go ahead and show you pictorially here.

Meaning that on the one hand we had dot, dot, dot, and we had a box in Pascal's triangle. And if you added it to the next element, then you got the one directly below. Right?

We had this going on. Okay? That was the Pascal's triangle rule that I stated by showing you earlier.

Then the rule we have now is going to be because n, j minus 1. 1 plus NJ. See, that's the next term as you go from left to right. The N say I'm in the Nth row. The J minus 1 to J tells me that these are terms that are right next to each other. That equals the N plus 1 row and the Jth term in it.

So that is the binomial coefficient version of this rule for row generation. So this is row N that I'm talking about here. And this one is rho n plus 1 here and here.

And this was Pascal's triangle, and this was the binomial coefficient triangle. And by showing these two rules are the same, which I'm about to prove by proving this statement over here, this starred statement, by proving that, I will show, therefore, the triangles are identical. And so I can interpret them either way, which will make a big difference for us when we're talking about the binomial theorem, which is coming up. Now here's the proof. And let me be frank about this.

This is an exercise in factorials. Exercise in factorials. So, you may have to listen to this more than once to catch all the factorial cancellations.

I'll try and show them all to you neatly, but this is a bit complicated with the factorials. n choose j minus 1 plus n choose j. Okay, this is... The left-hand side, here's the way I'm going to do this now.

I'm going to take this left-hand side and show it's equal to the right-hand side here underneath my head, the 2n plus 1j. I'm going to show this is equal to this. So I'm going to start out with the left and hope I end up with the right. That's what I start out with. Well, the first thing I can do is write out the definitions of both of these using factorials.

So the first one is n factorial over j minus 1 factorial. And then the difference between these two. that's n minus j minus 1. So I'm going to write that as n minus j and then plus 1 running the minus through. So n minus j plus 1. And that's factorial of course. Plus n factorial for the next one and then j factorial and n minus j factorial.

So all I've done is write down the definitions. Now what I want to do is combine these into a single fraction. In order to do that I've got to find the common denominator. So what I'm going to do is look at what I have underneath here and see if I can discern what that ought to be.

Now on this side over here, n minus j plus 1 is one number past n minus j. It's n minus j plus 1 after all. So this part here can be rewritten as n minus j plus 1 times n minus j factorial. I'm just taking the first number of the factorial out.

Likewise here, the j can be written as j times j minus 1 factorial. Now why did I bother doing that? I did that because I had a j minus 1 factorial over here, and over here I had an n minus j factorial.

And now with these observations, move my equal sign down a bit here, I can go ahead and write these out so that they look a little bit more the same. This one I will write out with these parts. So here's n factorial still on top.

j minus 1 factorial and then n minus j plus 1 the number times n minus j factorial. KELOLAND dot com. plus, writing this out, n factorial, and then I have j here, j minus 1 factorial, and n minus j factorial. Well, that's good. Now, these denominators are still not the same.

What's missing? Well, let's see. They both have j minus 1 factorial.

They both have n minus j factorial. This one has an n minus j plus 1 that this one does not. So, let's go ahead and give it to it.

So, multiply top and bottom by n minus j plus 1. Now this one on the right has a J that the first one does not. So we will multiply the top and bottom here by J. That will give us a common denominator here and that will take us to the next page and we can write these out as a single fraction now.

So our common denominator will be the j, the j minus 1 factorial, the n minus j plus 1, and the n minus j factorial. And on the top we have j times n factorial. factorial plus that n minus j plus 1 times n factorial.

And that's what happened on top. Remember what we had previously here. So the j times n factorial and here it's the n minus j plus 1 times n factorial.

So finally we're at this stage and now the simplification is much easier. I can recombine these two into a single factorial as I can do here. In the top I can simplify by seeing that n factorial is common and factor it out.

So n factorial times j plus n minus j plus 1. And the bottom becomes simply j factorial times n minus j plus 1 factorial. Well, the top is even more simple. simple.

Notice j and minus j cancel. So I'm left with, if these two add up to zero, I'm left with n plus 1. Well, n plus 1 times n factorial is n plus 1 factorial over j factorial. And this one I'm going to rewrite n minus j plus 1. I'm going to reorganize it as n plus 1 minus j factorial. So I just reversed the positions of the minus j and the 1. But now you immediately read this by definition.

This is n plus 1 j, the binomial coefficient. Because remember how we do this? n plus 1, n plus 1 factorial, j, j factorial.

And this is the difference, n plus 1 minus j factorial, which it is. So to take us back through this, what were we trying to show? We were trying to show that this object here equals this object here. And we have now done that. And that will be the result that we want to use.

That is the result from the theorem. Let me bring the theorem back so you remember it. That says that the Pascal triangle and the binomial coefficient triangle are the same. These are the same triangle because they have the same row generation rules.

And it is in the form of binomial coefficients that we will see this triangle again. And that will be coming up in the next segment, which is finally the binomial theorem. a major result.

And now we come to the major result which is the title of this page at the top of your screen, the Binomial Theorem. Finally, we get to this, the binomial theorem, how to expand a binomial, x plus a, to the nth power. This is a major theorem. It is a nice piece of work to end this unit. And we've got all the preparation now from having looked at the binomial coefficient symbol and Pascal's triangle.

So let's put it all together. Okay, first of all, on the binomial theorem, how to expand x plus a. To the end, what I've done is I've written out the first few powers. So we can detect what sort of pattern might be here. So I've got the powers from 0 up to 4 here.

And I've written them out in a suggestive form. It's a triangle, which should suggest something. So I've got 1. I've got x plus a, x squared, plus 2x, a plus a squared. And then the x cubed and the x plus a to the fourth row here. And then, of course, this continues forever.

Now, I don't know if you see anything here yet, but there are some patterns that will become familiar. And I think what I'm going to do is keep this and come back to it because I want to organize this differently. It's a little hard to see what's going on.

I like to organize it this way, and I'll put it down this way. Can we detect a pattern? Okay?

And there may be several, but let's see if we can pull all of these together. And you know, when we're done with this, I'll make an observation that I think you'll find interesting. X plus A to the 0. Okay?

Let me write this vertically the following way. Let me put the coefficient 1 here, and then put X to the 0, A to the 0. Okay, x plus a to the 0, of course, is 1. So I'm writing out 1 in a rather elaborate fashion. 1 and 1 and 1 in this fashion. Now, you'll see from the next two what I'm trying to do here and what the pattern is. Let's look at x plus a to the 1 power.

And let me write that out as follows. The coefficient first, 1, x to the 1 times a to the 0, and below it I'll write plus the next one whose coefficient is 1, x to the 0, a to the 1. Now, I'm going to write the coefficient first, 1, x to the 1 times a to the 0, and you know that a plus, x plus a to the 1 is just x plus a after all. And again, this looks more elaborate than it needs to be.

But as we get on, you'll see that all of this is necessary. This is x to the 1 again. That's a 1, that's a 1, so this is just x. There's a to the 1, so that's a, and these are just 1. So it is x plus a.

Okay? So there's the first one, and here's the second one. Now it begins to look more interesting.

x plus a squared. Let me write this out this way. Coefficient.

1, x squared, a to the 0, plus coefficient 2, x to the 1, a to the 1, plus coefficient 1 again, x to the 0, a squared. Now don't worry too much about the coefficients that I have boxed just yet, but I think you should observe a pattern for the x's and the a's. It's most evident in this last one, but it is true in the first two also.

In the last one, notice that the power that I'm raising the binomial is 0. binomial 2 is 2, the first power of x is 2, and it goes down to 0, 2, 1, 0. Notice that simultaneously the a's go from 0 up to 2. So they're going in reverse order. And we'll get to the coefficients later. Same thing happens here. 1 to 0, 0 to 1. And of course here they're both the same because there's only one place to go.

Now let me continue this pattern a little bit further so that you'll begin to see it. I'll do it for the cube and for the fourth power. Okay, x plus a cubed. This can be written as 1x cubed a to the 0 plus 3x squared a to the 1 plus 3x to the 1a squared plus 1x to the 0a cubed. Now, you may recall that this is exactly how it looks.

Notice that the x powers go down from 3 to 0, the a powers go up from 0 to 3. Nice pattern. You still may not recognize the numbers yet, but don't worry, I boxed them in. for a reason so we can come back and look at them.

Let me do one more just to set this in your mind. X plus A to the fourth power is equal to 1 times X to the fourth A to the zero. And apart from the coefficients, you should be able to guess all the coefficients.

all the x's and a's now. x cubed a to the 1, that's a 4 there, and a 6. x squared a squared plus another 4, and this is x to the 1 a cubed plus 1, and this is x to the 0 a to the 4th. And again, you notice the x powers go down from 4 to 0, and they go up from 0 to 4 for the a's.

And then there are these coefficients, and then this course continues on out to infinity, as many times as you want to take the binomial to a power. Well, let's write down what we've observed so far, and then go back and see if we can pick up what we can say about the coefficients. Observe. Well, Well, the first thing I observe is that each terms, I didn't mention this, but each terms powers of x and a add to n.

Now, did you notice that? Let's come back here. We noticed that the powers of x go down and the powers of a go up, but did you notice that the sum is always 3? Here, 3, 3, 3, and 3. Same thing here, add to 4, 4, 4, 4, and 4. So that's an observation we could have made also. So each term's powers of x and a add to n.

The x powers go from n down to 0. So that's down. The way I organized it, and the a powers go from 0 up to n. Okay, those are the observations about the powers of x and a.

What about the coefficients? Well, the coefficients have the following. pattern and let's see if we go back and pick those up. First of all we have a 1, a 1, 1, a 1, 2, 1 and then continuing here we had a 1, 3, 3, 1, a 1, 4, 6, 4, 1. If I wrote those out just as they appeared, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, and then 1, 4, 6, 4, 1, and continued out to infinity.

Do you happen to notice what that is? That is exactly what we called previously Pascal's triangle. Or if you like, the binomial coefficient triangle, and now you know why it's called the binomial coefficient symbol and the binomial coefficient triangle, because the numbers are binomial coefficients after all.

This is exactly... Now this is not the way we drew it, but you can see that the rows here, which are now columns, are the same. There's 1, there's 1, 1, there's 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1. In fact, this is the way that Pascal actually drew the triangle.

drew the triangle when he was working on it originally. So he must have been thinking something along the lines of the analysis we just did. OK, with all that preparation, we can finally write down the theorem that we've been trying to get to.

Theorem. This is the famous binomial theorem. Okay, just like that and it goes like this if X and A are in the real numbers. So those are the two numbers that are going to be part of the binomial. And N, for us, will be in the natural numbers only.

We'll talk about powers like 1, 2, 3, etc. Now, I did use the power 0 in my examples, and we could extend this to include 0, but we'll usually be looking at powers that are higher than 2 and 3 and 4 because those are the ones we really want to expand. And we don't know how to do those easily any other way.

All right. Then, here's what we get, x plus a to the n. What could this possibly be if you try to multiply it all out? Here's what it is. It is the summation of the n choose j's times x to the n minus j times a to the j, where the j will go from 0 to n.

Okay, so there'll be n term, there'll be a number of terms in this, actually n plus 1, and these are the way the individual terms will look. The coefficient here is from Pascal's triangle or the binomial coefficient triangle. This is the n, this is the power of x and notice that we start out with j equals zero. So the power of x starts out n and then it continually goes down as we saw.

A to the j starts out with j equals zero and will go up as we saw. And so this is the result that we've been trying to get to. And let me go ahead and write this summation out a bit so you'll see what we're talking about.

If I write out a few of these terms, I start out with n choose zero. x to the n, because in this case, a to the 0, of course, is 1. plus n choose 1, x to the n minus 1, a to the 1, plus dot, dot, dot, and then I'll give you what the last two terms look like, dot, dot, dot, plus n choose n minus 1, x to the 1, a to the n minus 1, plus the last one, which is n choose n, a to the n, and the x to the 0 is really there, but I'm not writing it because that's just equal to 1. Now I won't prove this. Okay, we've set the scene for this. We've analyzed quite a We've looked at the coefficients, the powers of x and a, and I think we've set up how this ought to go. We won't prove this, but we will talk about this in detail.

So let's go ahead and first of all rewrite that with maybe the boxes notation. Maybe that's helpful to have. x plus a to the n would be, now here I'm going to use the box for j. So it'll be n choose box. Times x to the n minus box times a to the box.

Where the box goes from 0 to n. So that's a nice way of seeing where that index number takes its positions. Put a box in there because that's unusual and it catches your eye. So there's a box here, a box here, a box here. That's where the index goes.

What's a typical term look like? This is sometimes a question you're asked. What does an individual term happen to look like? Well, the box plus first.

term. Remember we're starting with the box equals zero and we're going to n, so that means we have n plus one things added. So the n plus first term or the box plus first term will be this object up here.

n choose box. x, times x to the n minus box, times a to the box. There it is. That is the n plus box plus first term for the box varying from 0 to 1 to 2 up to n. Okay?

I've been saying n plus 1. I've been trying to avoid that. Say box plus 1 because that really does stand for the j. Okay, one more thing I want to recall before we do an example.

Recall the fact that we had about the binomial coefficients that I said would be useful, that n choose j is the same thing as n choose n minus j. Now we're going to use this in our examples coming up, so I wanted to make sure that was fresh in your mind. So a typical term looks like this, and this is a fact we will be using. Example, let's go ahead and expand a binomial that we would prefer not to write by hand.

X plus 2, say, to the fifth power. Now, yes, we can go ahead and expand that by hand, but that would be tedious. And as the powers get higher, it gets even worse.

So, let's write down that X plus 2 to the fifth by the theorem is the summation I'll 5 choose J. Now notice the way I'm writing this. 5 is my exponent here, so that's what's going to be the top of my binomial symbol.

The bottom is always J, so I do 5 choose J to remember that. And then I come back here and write J equals 0 to 5, and then I put in the rest of it. X to the 5 minus J times 2 to the J.

And in this case, if you write it all the way out, Let me put a line here for clarity on the screen. 5 choose 0, and we have x to the 5th, because x's are going to start high and go down, and the a's start at a to the 0 and go up, plus 5 choose 1. x to the fourth times 2 plus 5 choose 2 x cubed times 2 squared and let's come down here 5 choose 3 x squared times 2 cubed you see the X's are going down and the twos are going up 5 4 X to the 1 times 2 to the fourth plus finally 5, 5 times just plain 2 to the 5th because here is where x is equal to the 0 power. Okay? Advice.

Use that fact. That we had on the previous page. That I'll even write again that n choose j is equal to n choose n minus j.

Because what this does is it allows you to use the symmetry of Pascal's triangle or the binomial coefficient triangle. Which means. that 5, 2, 0, and 5, 5 are going to be the same number.

We happen to know that's one from our experience. But the point is they're the same. As we come in, these two are going to be the same. And then finally these two will be the same.

And if there were an an odd number, there'd be one in the middle that we'd have to calculate. But at least we can use all of this symmetry to make our work go a little bit more quickly. So writing some of that out, here 5 choose 0 is the same as 5 choose 5, and we already learned earlier that that was 1. And 5 choose 1 is the same thing as 5 choose 5 minus 1, which is 4. That was equal to 5, and this is Recall because we did this earlier. I showed you examples of that.

The only one I don't know is 5 choose 2, which is the same thing as 5 choose 3. So let's calculate it. That's 5 factorial. I'll do the first one. 2 factorial, 3 factorial. Once again, I'm going to take the bigger one, the 3 factorial, into the 5 factorial.

That will leave me with 5 times 4 on the top over 2 factorial, because the 3 factorial over 3 factorial, the one that's inside the 5 factorial, will cancel with this one to be 1. That's 20 over 2 factorial. 2 factorial is just 2, so that's 10. So finally... I can write down what I wanted to write down.

What is the expansion of x plus 2 to the 5th? Well, coming from the previous page, it's x to the 5th plus 5 times x to the 4th times 2 plus 10 times x cubed. times 2 squared plus 10 times x squared times 2 cubed plus 5 times x times 2 to the 4th plus 2 to the 5th. And we can go ahead and finish that off by multiplying all these numbers out.

out the 5 times the 2, the 10 times the 2 squared, 10 times 2 cubed, 5 times 2 to the fourth and multiplying out 2 to the fifth to get finally x to the fifth plus 10x to the fourth plus 40x cubed. cubed, plus 80x squared, plus another 80x, plus finally 32. And that is the expansion of x plus 2 to the 5th that the binomial theorem allows us to do. In order to do this, we needed to calculate, let me bring back the previous sheet here, when we started this.

We have to be able to calculate all the binomial coefficients that appear. Plus, we have to look at the powers, in this case the powers of 2 that appear, that might need to be calculated. It's time for you to try a problem.

Let me pose this one to you. Looks a little different from the one I just did, but I gave you a hint so I don't think it'll be too bad. Expand 2y minus 3 to the fourth power.

A little less power, so it's not quite as much work. And you have a binomial here, but the theorem talks about x plus a. So my hint is to read this as 2y plus a minus 3. So the 2y can act like the x would act, and the minus 3 will act like the a would act in the theorem.

So you go ahead and try to expand this to the bitter end, and I'll come back in a moment and show you what I did. All right, here we begin. We're looking for 2y minus 3 to the 4th power.

That's what we'd like to expand. Okay, well, simply applying the binomial theorem as it stands, this is going to be a summation of, and it'll be 4 choose j, and the j will go from 0 to 4 then, times the first object, which is 2y. to the 4 minus j power times the second object, which I realize is minus 3. So that's minus 3 to the j power.

Okay, that's just writing down what the theorem tells me to write down. All right, I know I need to do a lot of work here, so let me write out the entire expression just so I have it in front of me. 4, 2, 0, and then I'll have 2y to the fourth. Of course, the other part is to the zero, so I don't need to write it. Plus... 4 choose 1, 2y quantity cubed times minus 3 to the 1 plus 4 choose 2. And now I'm going down in powers, 2y squared minus 3 squared plus 4 choose 3, 2y to the 1 power, minus 3 is going up now, cubed.

Plus 4, choose 4. Minus 3 to the 4th. And, of course, here the 2y is to the 0 power. And let me run a line through here again so it will be a little easier for you to see. Okay?

That is what that is totally multiplied out. Now you see what work I have to do. to figure all the binomial coefficients and then I have to do other multiplication.

If 2y is cubed, I have to have 2 cubed times y cubed. And then all these various powers of minus 3 need to be dealt with. So, let's start with that. 4 choose 0 is the same thing as 4 choose 4 and that's 1, we knew that. And then 4 choose 1 is the same thing as 4 choose 3. We also knew that was 4. The middle one, the 4 choose 2, we don't know, so let's calculate it.

4 factorial over 2 factorial times 2 factorial, because this 2 is 4 minus 2. But notice that 2 factorial and 2 factorial are both equal to 2. And 2 times 2 is 4. And 4 will cancel with one of these up there, leaving me with plain 3 factorial. But 3 factorial we know is 6. So I now have that binomial coefficient finished. So, I would want to go further and multiply all of these minus 3's, pull the powers of 2 out. And so with that in mind, knowing that you've done this, I can now write out that 2y minus 3 to the fourth power is equal to 16, that's the 2 to the fourth, y to the fourth, plus 4 times 8 times minus 3y cubed, and I'll need to combine those.

plus 6 times 4 times 9, I'll combine those, times y squared, plus 4 times 2 times minus 27 times y, plus 81 because that's 3 to the 4th, and minus 3 to the 4th is the same thing, of course. So finally I have 16y to the 4th minus 96y cubed plus 216y squared minus 216y plus 81. And there you have it, the expansion that you were looking for. And I hope that's what you got and that it wasn't too difficult. Further, I want to pose the following question, because you are very often asked this kind of question in problems.

Which term... contains a particular power of k say x to the k more often someone will ask you a specific question like in the expansion of like the one we did 2y minus 3 to the fourth which term contains the power y cubed okay find what that term is find the coefficient etc Well, the answer is to recall what the typical term looks like and then translate it into what we need. The typical term of the binomial expansion, if you recall, was n choose box, remembering my boxes, times n minus box, x to the n minus box, times a to the box. Now what do we want here? We want the power of x to be k.

So that means we want n minus box. to be our K. Well, then what is the box going to be?

Well, it's clear that N minus K then is equal to the box. That's just a tad of algebra. So what do we have then?

We know that the, what we're trying to do here is is get the term containing x to the k. So the term containing x to the k is now going to be equal to, let's see, putting n minus k in for the box, I have n, choose n minus k, and the n minus box is k, so this is times x to the k, which we were expecting, times a to the n minus k. And there it is.

If you're looking for the term containing x to the k, there is that term. Now this can be handy if you have a specific question in front of you. For example, let me show you how this works. Here's an example.

Find the coefficient, the numerical coefficient, find the coefficient of y to the 8th power in the expansion of 2y plus 3 to the 10th power. Now the idea here is that you do not want to have to expand this to find out that coefficient. You want to do as little work as possible.

So, first note that the term containing y to the 8th, which is what we're looking for, is the same as that containing... Well, since we have the term 2y there, 2y to the eighth, of course. It's whatever contains that. So that's an observation. Then from above, the fact above that we just computed, from the fact above, with k equal 8 here and n equal 10, we have that n over n choose k.

times x to the k times a to the n minus k, okay, is equal to 10 choose 2 times 2y to the 8th times 3 squared. And we know that we want the numerical coefficient of y to the 8th here, so we need to work some things out. So let's first of all deal with 10 choose 2. By definition, that's 10 factorial over 2 factorial, 8 factorial, because 8 is 10 minus 2. And then dividing the 8 factorial into the 10 factorial, I'm left over on the top with 10 times 9 over 2, because that's all the 2 factorial is. That's 90 over 2. That's just 45. We'll go to the next page and pick out the other numbers we need here.

We need 2 to the 8th. 2 to the 8th is just 256, and finally, the 3 squared, of course, is 9. So, we end up with the final coefficient is going to be 45 times 256. times 9 all times y to the eighth. That comes out to be 103,680. That's the coefficient we were looking for right there. 103,680.

That's the coefficient that is the coefficient of y to the eighth in the expansion of the binomial that was given. So sometimes questions like that are asked, and I wanted to show you at least one example. But I wanted to leave you with a puzzle, and I'll come back and show you how to attack this.

You can attack this using the binomial theorem, which is kind of nice. And I like this. This is a pleasant puzzle.

The question is, which is greater? 1,000 to the 1,000 power, or 1,001, a little bigger, to the 999 power, a little smaller? Which one of these is bigger? Now my hints are the binomial theorem helps, so you might consider using the binomial theorem. There's a new fact that I'm adding right now that I think will help you through this.

It helped me in doing it. The fact is that n choose r is always less than n plus 1 to the i. our power. So just jot that down and use it if it seems appropriate.

The final hint is that your calculator probably won't help. These numbers are just too big for the calculator to distinguish them. It may even say that they are the same. They are not the same.

One is actually bigger than the other and I won't tell you which. So why don't you go ahead and try that and I'll come back in a moment with my solution. Well, has the suspense been building? Let's see what the answer is going to be.

As I said, I'm going to use the binomial theorem here. So by the binomial theorem... It will allow me to break this up in a very handy way. I will take 1,001 to the 999 power and rewrite that as 1,000 plus 1 to the 999th power. Now, of course, it's a binomial, so I can use the binomial theorem.

And to be specific, I'm talking about the case where n, the power, is 999, the x value is 1,000, and the a... the other part of the binomial is just one here. Okay, what would the binomial theorem tell us? Well, it tells us this is equal to a sum of these coefficients, 999 choose j, and then the j will go from 0 to 999, times, let's see, the first term is 1,000 to the 999 minus j, times the second term, which is just 1, to the j power. Now, of course, 1 to the j...

It'll just be one, so we'll drop this out from the subsequent writings of this. But the very first thing I'm going to use is that fact that I mentioned on the previous page so that I can get this inequality. Let me remind you what the fact was.

The fact was this, that n choose r is always less than n plus 1 to the r. Now, there may be other ways to do this problem, but this is certainly a very easy way that I saw. So I'm going to use that on the 999. choose J here and I'm going to say it's less than this object so this will make this less than the summation from J equals 0 to 999 of In place of this object here, this 999 choose j, I will put the other side of the inequality, the n plus 1 to the r.

Now in this case, if n is 999, then this is 1000, and the r is j, so that's the jth power. So I have 1000 to the j times 1000 to the 999 minus j, and of course I won't write that 1 as I said I wouldn't. But now look what I have here. I have...

something that can easily be combined into a single power. A power 1,000 to the j plus 999 minus j, which of course is just 999. So let me put that on this other page here. And just for the record, let me remember that my inequality is going that way, so I don't forget from page to page.

And now that's going to be equal to the summation of, as I said, 1,000 to the 999, where j goes from 0. to 999. Now that is a constant. There is no J, the index letter, in there. How many terms are there? How many times do we add this constant in this context? Well, the J goes from 0 to 999. That's 1,000 terms.

So let me write that down. 1,000 terms. And if that's 1,000 terms, then the inside is simply going to be multiplied by 1,000.

So 1,000 times 1,000 to the 999. And of course, that immediately simplifies to 1,000 to the 1,000 power. So that was the other side of our inequality we were hoping to get to. So just to put it all together here, we've discovered that 1,001 to the 999, remembering the direction of my inequality, is less than 1,000 to the 1,000 power.

So, I hope you found that enjoyable and an interesting little problem. That will be the end of Unit 7 for us, and from here you can go on to do a variety of other things.

Transcript for:Understanding the Binomial Theorem

Transcript for:
Understanding the Binomial Theorem