Welcome back, Algebra 2 Wonderkind! Today we're going to be talking about sequences. So, for example, maybe we start with one creepy Kelly head, floating head. And then we double it, and then we double it, and we double it, and we double it, and we double it until it's just too much creepiness to take.
So in other words, we have a sequence here, and we have something that we're doing to the term previous to it to get to the next term. Let's take a look here. The algebra was give a huge presentation. Sully sets up all the chairs.
The first row has 10 chairs. So row one, 10 chairs. Each row after that has two more chairs. So let's see. Row two, I would add two here and I'd get 12. Row three, I would add two here and I would get 14. Row four, I would add two to the previous one and get 16. So I have completed the table.
I could do row five would be 18. Every time I'm adding the same number over and over again. So let's talk a little bit about that. What is happening here?
So this is called a sequence. A sequence is an ordered list of numbers. In other words, it does matter what order we are in. Our first term is 10. Our second term is 12. Our third term is 14. They relate to each other.
The fourth term is 16. The order of these things matters. All right. A recursive rule is a rule to define the term in relation to its previous terms.
And we use it to find the next term. Now, we kind of did this. All right.
We've done this before in Algebra 1. We kind of did a recursive rule, right? We had our previous term. So let's say I had a to the n. My previous term would be a to the n minus 1, and I added 2 to it. Correct?
Now we have to understand also that we it makes a difference what our first term is our first term in this case was 10 the first row had 10 and that's awesome recursive rules are very cool It helps you get to the next term now Honestly getting to the next term at this point in your mathematical period probably isn't too difficult. All right, so Let's talk about the next one. The next one is called an explicit rule And that is a rule that is using the location of the term.
So maybe I want to know how to get from the fourth term directly. I don't need to go, I don't want to know how to go from the third to the fourth. I want to go directly to the fourth term. And it can be used to identify any term.
So in other words, we're going to find the nth term. So let's go back here and look and see how we can find the nth term. Let's rewrite this one. So, well, let's do this one first. How did I get here?
I started with 10 and then I added 2 to that, right? This one I started with 10 here and then I added 2 twice, right? This one I had 10 and I started, I added 2 three times. Okay, let's see if there's an easier or better way to write that. So this is like repeatedly multiplying something, right?
So this is like 10 and I had 2. I multiplied it 3 times because I added it 3 times. Let's see this one. I had 10 and I added 2 twice, right? This one I had 10 and I added 2 one time.
Ooh, let's see if we can do the first one. I have 10. How many times do I add 2? I add 0 times.
So 10. I multiply by 2. 0 times. So now we want to find the nth row. Let's take a look here.
How can I find the nth row? What's the formula for the nth row? Well, there's a couple things in this situation that absolutely are not changing. The 10 is not changing, so that's going to be part of my formula. The plus 2 is not changing.
In fact, the only thing that changes is the number I multiply by. correct? So we want to find this and this is going to be our n. We want to see how are we going to put n into this formula.
Let's just take a guess here. Let's try one. Could I make this formula 10 plus 2 times n? Let's see, does it work for step or row term 1? 10 plus 2 times 1. It doesn't.
I needed to subtract 1 from that to get 0. So if my, alright, let's try row 2, nth term 2. If I subtract 1 from it, then I get 1. That's what I want to multiply by. Here, I'm at rho n is 3. If I subtract 1... I have 2, which is what I want to multiply by. Here, I had 3, but I wanted to subtract 1 to get to, excuse me, I was at 4, row 4, subtract 1, I get to n is 3. So really, if I want to write row n, I want to subtract 1. So our explicit formula would be 10 plus 2 times n minus 1. And the great thing about that is now we have a number of chairs.
We have a formula. We can find any. If I want to find the 10th row, I would do 10 plus 2 times 10 minus 1 because n is the number of rows, row 10. 10 plus 10 minus 1 is 9. 2 times 9 is 18 plus 10 is 28 chairs. So it really helps us out quite a bit if we have that explicit formula.
Okay, so that first sequence we were talking about was actually called an arithmetic sequence. And an arithmetic sequence is a sequence in which consecutive terms from one term to the next have a common difference. In other words, we are going to be adding the common difference every time.
And our common difference in that first example was two. I added two chairs every single time, right? So what I'd like you to do right now, I'd like you to pause the video. I'd like you to try to answer these questions here, okay?
Here's my table. N is one. The value of one, the value of one, this is just member function notation. This is y. So 80 would be...
f of 1 equals 80, right? The value of the function at 1 is 80. So why is this arithmetic? What's the common difference? And try and find an explicit formula.
So find the nth, the nth term. So go ahead right now and pause the video. Okay, let's take a look.
Why is this an arithmetic sequence? Well, we are every time... I'm subtracting 6. Now is subtracting 6 adding a common difference? Well no, it's subtracting, but really it is the same thing.
What if I'm adding negative 6 each time? So therefore, I am in fact have an arithmetic sequence. I just have to think about it instead of subtracting is adding negative 6. So what was our common difference?
Our common difference was Negative 6. All right, let's take a look here. How can I get this one? I started with 80. That's good. To get to this one, I did 80 minus 6. To get to the next one, I did 80 minus 6 minus 6, right?
Down here, 80 minus 6 minus 6 minus 6. So let's take another look here. Give us some room. All right. So, this third one, I like this one, I'm going to start with.
This is my initial value, 80. I'm repeatedly adding negative 6, right? So I'm repeatedly adding negative 6. How many times did I repeatedly add it? One, two, three times.
All right. Down here, I have 80. I started with 80, and I repeatedly added negative 6 twice. Up here. I started with 80 and I repeatedly added negative 6 one time. This one, I have 80 and I repeatedly add negative 6 how many times?
No times, right? So let's see if we can come up with a formula. What doesn't change?
Well, the 80 doesn't change. The plus negative 6, or we could just say minus 6 doesn't change, right? This changes every time. Now, how does 3 relate to 4? Well, it's 1 less, right?
3 is 1 less than 4. n is 4, 3 is 4 minus 1. How do I get to 2? 1 less. 3 minus 2, or minus 1 is 2. 2 minus 1 is 1. 1 minus 1 is 0. n minus 1 is n minus 1. And there we have our explicit formula, all right? So let's copy and paste that real quick. So we have our explicit formula here.
80, whoops, 80 minus. 6 times n minus 1. Can I copy and paste this? What do you guys think? Pretty good, Sullivan. All right, so here's our formula.
All right, so now we want to know what is the 10th term. So I want to find the value of the 10th term. All right, so my y value, the value of my function, I want the 10th term.
So I come over here, and I do 80 minus 6. Instead of n this time, I'm putting 10. 10 minus 1. So this is 80 minus 9 times 6. That was backwards, sorry. 80 minus 54. So the value of my function on the 10th term is 80 minus 54, which is 26. So the 10th term is 26. Good. Now this says, what is the term number?
So I don't want no n. It has a value of negative 10. So in this case, I'm saying f of n equals negative 10. So here we go. Negative this whole thing, negative 10 equals my formula, 80 minus 6 times n minus 1. All right, let's do our order of operations backwards. We've got to do the opposite of add 80. We subtract 80. We get negative 90. I could distribute here, that's true, but this is multiply. The opposite of multiply is divide both sides.
So I have 15 equals n minus 1, and then add 1, and I get 16. So my 16th term is going to be a value of negative 10. f of 16 is going to be negative 10. Okay? Alright, so the question is, is there a general explicit formula for an arithmetic sequence? So...
If I had an arithmetic sequence, we have two over here, right? We had this one, 80 minus 6 times n minus 1. And we had the other one, 10 plus 2 times n minus 1. Do we have a general format? Well, if I look at this one, 10 was my original number right here, correct?
And I added the common difference times n minus 1. So let's take a look over here. I have 80, which is my initial, my first term. Then I subtracted the common difference add negative 6 times n minus 1 so it appears to me as though I'm gonna do my first term whatever the value my first term is plus my common difference so here we go in Fact that's what it is so we have the value of the function Equals a plus D times n minus 1 a is just the first value so when n equals 1 D is the common difference and N is the term you want to find. So if I had a formula and it was like if I had a sequence and it was like this the first term is 4 so my formula for this my nth term would be my initial value plus my common difference here I'm going up 3 every time so plus 3 every time times n minus 1. It's that easy to find.
Initial value, common difference times n minus 1. All right, so here we have our next one. Mr. Kelly tells his fifth kid that the first quarter he has straight A's, he'll give him $5 at the end of the first quarter. So then every quarter after that with straight A's, he'll double the previous amount.
So the second quarter, he'll get how much? Double it, so I would multiply by 2, so this would be 10. After the next quarter, double it, multiply by 2. That would be 20. Nice. After the fourth quarter, double it, and he would get $40 for the fourth quarter if he got straight A's. That's pretty nice, Mr. Kelly. Nice job.
So the question is, how much money does he have after eight quarters? Well, we could keep going right here, right? So I got 5, 10, 20, 40. So this 1, 2, 3, 4. The fifth quarter would be 80, 6. 7th, and 8th quarter.
So if he does 8 quarters in a row of straight A's, he'll get $640 for that quarter. That's pretty awesome. All right? All right.
What's an explicit formula for this one? So we have our recursive thing here. We're multiplying every time by the same number.
So let's rewrite this first one here. So I had, how do I get to 10? I had five dollars and then I doubled it.
I multiplied by two. Here I had Five dollars, I multiplied it by two once, two twice. So I have five times two times two. Here I had five times two times two times two. All right, let's rewrite these a little bit.
Did I multiply this by anything? Ooh, this is gonna be a good question. I multiplied it by what? One, really, right? So we're gonna think about that one in a second.
All right, let's see here. So I have five. What's a shortcut for multiplying the same number three times? Well, that would be 2 to the third power, right? Exponents are repeated addition, or multiplication, excuse me.
So 2 times 2, this would be the same thing as 5 times 2 to the second. 5 times 2, this would be 5 times 2 to the first. Let's look at our rule of 3, 2, 1. So this is going to be 5 times 2 to the 0. And you may remember, anything to the 0 power is 1, which is what we got here. So can we come up with our formula? Let's see here.
What doesn't change? 5 stays the same. 2 stays the same. So this changes.
3, 2, 1, 0. So let's relate this. Our fourth term... We wrote 3. That would be 1 less, right? Fourth term, 1 less would be 3. Our third term, we had 2, 1 less.
Our second term, we had 1, 1 less. Our first one, we had 0, 1 less. So if I'm at the nth term, I'm going to have 1 less.
And there you have it. Alright? This is not an arithmetic sequence.
This is called a geometric sequence. And a geometric sequence is a sequence formed by multiplying by a common ratio. So, our common ratio in the last one was we doubled it, we multiplied by 2. Now, before we talked about subtracting being the same thing, let's understand something. If I am dividing by 2, that is the same as multiplying by the reciprocal.
So if I had a formula that was doing 32, 16, 8, 4, my common difference would not be divide by 2. My common difference would be multiplied by 1. All right. What I'd like you to do right now is pause the video, and I'd like you to try to find and answer these questions right now. All right. So is this a geometric sequence?
Let's see. I multiply 2. 0.25 times 2 is 0.5 times 2 is 1 times 2 is 2. So why is this a geometric sequence? I am repeatedly multiplying. Boom. What is that common ratio?
The common ratio, what am I multiplying by every time? I'm multiplying by 2. Alright, let's see. What's an explicit formula? Here, this was my initial value, 0.25, times 2, 3 times.
1, 2, 3. This was my initial value, times 2, twice. This was my initial value, times 2, once. And this was just my initial value, right? Alright, so let's see, what doesn't change? Well, let's rewrite this a little bit.
So this is going to be... that this first one is going to be 0.25 times 2 to the first. The next one would be 0.25 times 2 to the second.
And the third one would be 0.25 times 2 to the third. So if I want the nth term in my explicit value formula, all right, I want the nth term. Let's see what I need to do. Well, my initial value.
my common ratio, right? And now this was the fourth term, I had an exponent of 3, that's 1 less. Third term, I had an exponent of 2, 1 less.
Second term, I had an exponent of 1, 1 less. And in fact, this would have been 0.25 times 2 to the 0, because anything to the 0 power is 1, so it looks like it's going to be to the n minus 1, all right? So we have 0.25 times two to the n minus one. Good. Let's use this formula now and answer some much cooler questions, right?
Alright, so we have our formula here. So let's see, what is the 10th term of this formula? So how am I going to figure out the 10th term?
Well, I know my n is 10. n is 10. So I want the value of my function at 10. Let's see what we got. 0.25 times 2 to the n minus 1, so that's 10 minus 1. That's 0.25 times 2 to the 9th. And honestly, that's calculator work right from there. Alright, I could do two to the ninth and then multiply it, but honestly, I just like putting all that stuff in at one time and I get 128. So the tenth term is 128. These are exponentials, right?
It means they're going up much faster than our arithmetic. Our arithmetic were linear. They were going up the same rate at all the time. These grow or decay much faster.
What term has a value of 8192? So in this case... My y value, my term value, I want 8192 to equal 0.25 times 2 to the n minus 1. This is going to take a little bit more work. So let's first of all divide both sides by 0.25.
All right, that gives us 32,768 equals 2 to the n minus 1. Now the trick is we have to remember how to solve and get these. I'm going to use logarithms. And I remember that if I do the log of this base, it's going to cancel. So I'm going to do log of base 2 of 32,768 on one side and do the same log on the other side because what I do to one side I have to do to the other. So now this tells me they cancel.
So what's left on this side? N minus 1. And remember, you can use your math button on your calculator and log base to calculate that. All right. And when you did calculate that, you would have gotten 15. So the log base 2 of 32,768 is 15. Add 1 to both sides, and we get 16. So f of 16 would give us 8,192. Very nice.
Wouldn't it be nice to know what a generic or general formula for this is? I think it would be nice. In fact, that's my next question. So let's take a look back here.
This formula, I had the initial value, 0.25, times my common ratio to the n minus 1. Let's look at the first one. I had my initial value, 5, times my common ratio, 2, to the n minus 1. Ooh, interesting. So our general geometric formula is going to be a, our first or initial value, times r, the common ratio to the n minus 1. Remember, n is the term you want to find. All right.
So let's take a look here. If I had this very, very, very simple formula right here, I could find it. It would be f of n equals my initial value, 1. times my common ratio.
What am I multiplying every time here by? 3 to the n-1 power. And there you have it. Okay, I'd like you to pause the video right now and try this one all on your own.
Alright, so let's take a look here. What do we have? So what is the explicit formula for this sequence?
Well, I found this to be multiplying by 0.5. Now, I'm going to be honest with you. I didn't recognize right off the bat that this was half.
I might have recognized this one. But so what I did was I went back and I divided. I know I need to multiply going this way. So I went backwards and divided.
So I did 1280 divided by 2560 in my calculator and it gave me one half. So that's how I know that was that. And I checked it each time to make sure that when I divided by the later one, I got that. So you can do that.
If you're not sure, you can always go backwards. Remember, if you're going backwards, you're going to do the opposite. Sometimes you can just see and you know what number you have to multiply though.
So over here, what's the explicit formula? So this is my initial value times my common ratio to the n minus 1. All right the 12th term so I put 25 60 times 1 half to the 12 minus 1 now a common theme I want you to understand is you cannot do 25 60 times 1 half because of our order of operations You have to do the exponent first. So I have to do 1 half times 11 first. Alright, and then multiply by 2560. Or you could put all of this in your calculator at once, as long as you don't put it in as 2560 times 1 half, all of that to the 11th.
If you group the exponent with the 1 half, then you'll be fine. Alright, when I did that, I got 1.25. Down here, what term has a value of 10? So I want to know, when will my... the sequence equal 10, right?
2560 times a half to the n minus one. I divided both sides by 2560. I recognize that if I took the base of this logarithm wise, log of one half, all right, it would cancel on this side and I'd get n minus one. The log one half of 10 to the 25, 10 divided by 2,560 is eight.
Eight equals n minus one. Add one and I got nine. So there you have it.
All right. Hopefully this isn't too troublesome for you. All right. We've been doing logarithms this year.
We've been doing sequences. It's not too terrible, hopefully, for you all. All right.
Remember to be nice to each other and be the change in the world that you want to see. Till next time. Peace out.