what's the difference between a sequence in a series well a sequence is a list and a list is separated by commas whereas a series is a sum you're adding up all the terms in that series that's you're adding them so some but first let's talk about the notation so each one of these terms you could call n equals 1 N equals 2 N equals 3 N equals 4 etc that tells you what turn that you're on where as a sub 1 sub means like subscript a sub 2 a sub 3 that tells you the value of that particular term so what we're going to do now is we're going to figure out how do we find out a specific term in this sequence well first of all the question is is it arithmetic or as a geometric now when we talk about arithmetic sequences we're talking about are we adding the same thing each time to get to the next term if it's geometric then we're multiplying by the same thing each time to get to the next term so here what are we doing to get to the next term it looks like we're adding to adding to adding to adding to they call that D equals 2 that's the common difference so D for difference like if you subtract 5 minus 3 you get 2 7 minus 5 you get 2 but say we wanted to find out what this sixth term is what we would do is we would start at 3 and we would add 2 once twice three times four times five times so why do we add to 5 times how come not six well because we were already at the first term three okay so we only had to add that difference not six times about 5 times and that's where our formula comes from here for arithmetic sequences you take the first term a sub one plus the common difference okay here D which is 2 times n minus 1 and is whatever term you want so in this case we want to term 6 that's 6 minus 1 which is 5 and you got it so this would be a sub 6 equals 10 plus 3 which is 13 which we we knew that but say you wanted to find like a term that was maybe like the 20th term well instead of having to add to add 2 at 2 until you got to the 20th term we could write a formula or a rule to find that so let's go ahead and do that so a sub n it a sub 1 which is 3 plus and minus 1 times D D is our common difference let's go ahead and simplify this by distributing the 2 so we get 2 n minus 2 plus 3 and let's combine like terms so 2n plus 1 because negative 2 plus 3 and so now we have a formula or a rule for finding any term if we want to find that 20th term we're just going to put in 22 times 20 is 40 plus 1 is 41 and it takes you right to that term now another way to write these rules is instead of writing an explicit formula formula it takes you right to that particular term you can write what it's called a recursive formula and the recursive formula looks something like this let me spec and show you you start off with telling the person what's the first term in the sequence in this case it's 3 so a sub 1 meaning the value of the first term is 3 and then you give them a rule or a formula for finding the next term now there's a couple different ways to do this one way to do it is to refer to the previous term so a sub n minus 1 that means one term before the term that you're on that term that you want to find and then we're just going to add a two so for example if we want to find the sixth term a sub six we would go to the six minus one which is the a sub five term here the value of the fifth term and we would just add two and we got it now the other way I was telling you that you could write this is instead of saying a sub n you could say a sub n plus one so that means you want to go to the the next term the one after the nth term and so then you would take a sub n that's the value of the term that you're on and you would add to either one of these is a fine whichever one you prefer now say for example you wanted to add up all these terms will say up to the 20th term how would you do that well let me illustrate here say you wanted to add all the numbers from one to a hundred well that would take you a while right but look at this pattern so if you take one plus a hundred you get a hundred in 1/2 plus 99 you get a hundred and one three plus 98 you get a hundred and one how many hundred and ones do you think that we could make you know by adding the numbers from 1 to 100 well if you said 50 you're right because you're pairing them up two at a time right so what we have here is we have there's 50 100 in once so if we generalize now and write this as a formula what we did is we took the number of terms which was 100 divided by 2 because we're pairing them up two at a time and then what we did is we took the first term a sub 1 which is 1 plus the nth term okay which is the last term which is a hundred and that's how we got our hundred and one so that's where this formula is coming from right here it's the number of terms divided by 2 times the sum of the first term in the last term okay so ban just means the the nth term of the last term so going back to this problem over here if we wanted to add up the sum of the first 40 terms ok S sub 40 we know that there's 40 terms ok we're dividing that by 2 because we're gonna pair them up two at a time and the first term we said was three and the 40th term let's see what the 40th term is we're going to put 40 in four here so that's 2 times 40 which is 80 plus one which is 81 and so now we have 20 times 84 let's see 10 times 84 is 840 so it's 840 double that's 1680 and that's the sum of the first 40 terms let's take a look at this sequence now we've got 5 15 45 135 dot dot dot what are we doing to get from this term to this next term to the next term to the next term what here you can see we're actually multiplying by 3 and this thing that we're multiplying by each time this is called the ratio so our four ratio and if you're having trouble finding out what that ratio is just take the number divided by the one before it same thing here 45 divided by the number before it and you can see we're getting 3 3 3 3 now if it looks like you're dividing remember division is really like multiplying by the reciprocal so if it looks like you were dividing by 2 that's like multiplying by 1/2 same thing with the arithmetic ones if it looks like you're subtracting it's really like adding a negative number so that's important to keep in mind so arithmetic you're adding the same thing geometric your mult by the same quantity but now let's say we wanted to get an idea about how to find the value of this fifth term here well you can see we're starting at five and we're multiplying by three how many times one two three four now notice that we're on the fifth term but we only multiply by three four times so that's where our formula comes from what you do is you start at the first term you multiply by the ratio n minus one time so that's the N minus one power so if we were to write this as a rule or formula it'd be a sub N equals a sub one times R to the N minus one our first term we said was five per ratio is 3 to the n minus one power and that's your rule or your formula now if you wanted to write it in a recursive manner you would do the same thing you would let the person know what term you're starting at which is 5 and then you'd give them a rule for finding that next term next term next term now recursive formulas are good for computers you know computer can do all these calculations quite quickly and you can write it like this where you're taking the N minus 1 term and you're multiplying by 3 so if we wanted to find the fifth term a sub 5 we're taking the a sub 5 minus 1 which is the a sub 4th term okay which is this 135 and then we're multiplying by 3 now if you wanted to find the 10th term you might not want to keep multiplying by 3 you know a bunch of time so you could just put in 10 for n and find the value of that tenth term now let's talk about how do we find the sum or switching from sequences to series now how would I find the sum let's just say the first 10 terms well of course we could write them all out and then add them up but if it was like 20 terms or 100 terms that would take too long so let's go ahead and figure out the sum of the first 10 terms we're going to be using this formula right here and the first term we know is 5 so that's a sub 1 we have 1 minus the ratio which is 3 to the 10th power because we have 10 terms all divided by 1 minus the ratio 1 minus 3 now when you work with this formula you want to be careful some students mistakenly will write 1 minus 3 oh that's negative 2 to the 10th power but you want to follow the order of operations you want to do the exponents first and then the subtraction okay sir with me so let me go to the calculator on this one and I'm just going to do it here you can do it in steps if you want so I'm going to do three to the tenth first I'm going to do one minus that answer I'm going to multiply by five and then divide by negative two and that comes out to one hundred forty-seven thousand six hundred and twenty and that's the sum of the first 10 terms now notice how this is such a large number and it's because every time you multiply by three you know you're tripling tripling tripling is growing exponentially so that's why we're getting into quite a large number quite quickly now another type of geometric series we want to look at are these ones here where you have it an infinite number of terms see the dot dot dot so it continues on and on and on as opposed to a finite set of terms so this is the form we're going to use for finite this is the formula use for infinite geometric series but the important thing here is is that when you multiply by that ratio that ratio has to be in between negative 1 and 1 so what happens you can see it I'm multiplying by 1/2 that's 50 1/2 that's 25 12.5 what is this number getting closer and closer to zero so that means that it's converging okay so it's approaching zero and we can sum all those terms up now if we were multiplying by two each time that would be what we call divergent and it would be going to gradually to infinity if we try to add all those terms up of course there would be an infinite sum so we wouldn't be able to actually get an exact you know value for that sum so here what we're gonna do now is we're gonna use our formula we're gonna say the sum of an infinite geometric series it's the first term which is 100 over 1 minus the ratio we're multiplying by 1/2 each time to get to the next term so this comes out to your 100 divided by 1/2 which equals 200 remember when you divide by a fraction it's like multiplying by the reciprocal again sometimes you know you may have a problem where it looks like it's getting smaller right but it might actually be diverging like for example here's a little trick sometimes that teachers might put on your test it might be something like this like 8 plus 6 plus 4 plus 2 and you might say oh it's getting closer to zero but what happens because this is arithmetic you actually just flew right by zero alright so you keep subtracting two and eventually this is going to go to negative infinity you can't actually find that some it's going to go to negative infinity right so it has to be geometric and the ratio has to between be between negative one and one for it to converge and for you to find the sum let's look at a couple more challenging problems so write or rule for the arithmetic sequence when you're given the third term is seven and the fifth term is 13 so how do we do something like that well let's go over here to our arithmetic explicit formula now let's take this term here we know that the value of the third term is seven so we're gonna call this seven the first term we don't actually know and the third term okay that's N equals three 3 minus 1 is 2 and the difference we also don't know let's take this term here the fifth term is 13 so same thing the value of that term is 13 the first term we don't know but 5 which is the term that we run 5 minus 1 is 4 times the difference which we don't know so what we have is a system of equations where we have two variables and two equations so we can solve for a 1 and D so what we're going to do is let's go ahead and subtract these equations we're gonna use the elimination method so the a ones would cancel out 7 minus 13 is negative 6 2 D minus 4 D is negative 2 D and if we divide both sides by negative 2 you can see that D is equal to positive 3 now if we put 3 back into either one of these equations I'll just put it into this top one here we have 7 equals a 1 plus 2 times 3 which is 6 and if we subtract 6 from both sides you can see that a 1 is equal to 1 so now if we put it all back together we can write a rule for this sequence we have a sub N equals a sub 1 which is 1 plus and minus 1 times D which is 3 and then I like to just kind of simplify this down a little bit by distributing and combine like terms we have 3n minus 2 equals a sub n now you could test it out if you want you can put 5 in for example 15 minus 2 that's 13 you could put 3 in 3 times 3 is 9 minus 2 is 7 you know checks out and that's our formula for our arithmetic sequence but for this one it says write a rule for the geometric sequence given two terms how would you do that one well same thing we're going to go to our geometric explicit formula the value of the second term is six and we don't know what the first term is and we don't know what the ratio is but we know that second term so two minus one and is equal to two so this is one over here we have 162 equals we don't know the first term we don't know the ratio but 5 minus 1 is 4 so now we have two variables the first term and the ratio and we have two equations we can solve that system what I'm going to do here is I'm going to divide both sides by R ok so now you can see that a one equals six divided by R we can put that in place of a 1 in the second equation so we have one sixty two equals six divided by r times r to the fourth now this is really like R to the fourth over one because anything can be written as a fraction and we can do some cross reducing one of these R's cancels with one of these R's leaving our cubed if we divide both sides by six let me go to the calculator on that one that comes out to 27 equals R cubed you take the cube root of both sides and you can see we're getting our equals three now if we put three back in to either one of these equations I'm going to put it over here we have six divided by three equals a sub 1 a sub 1 equals 2 so we have the first term in the ratio we can write our our rule or our formula for any term in the geometric sequence and if you want to check it out you can always say I'll plug in 2 for example 2 minus 1 is 1 3 to the first is 3 times 2 gives you 6 and same thing with the next term let's take a look at this summation notation so see this Greek letter Sigma here it kind of looks like an e that represents the sum and what you do is you take this number here on the bottom this is called the index and you would put it in so if I put one in I get 3 times 1 is 3 plus 2 is 5 ok that's our first term now if I put 2 in 3 times 2 is 6 plus 2 is if I put three in that's 9 plus 2 is 11 plus dot dot dot you work your way from 1 all the way up to 10 you know in order sequentially like that so if we put 10 and we get 30 plus 2 which is 32 and what the Sigma represents again is the sum or summation it's a series and we're adding up all these terms but what you can notice here is that see how are adding three three three this is an arithmetic series so to find the sum we're going to go over here to our arithmetic series formula we're gonna take n which is a number of terms ten divided by two the first term a sub one is five plus the last term which is 32 and we get five times 37 which is 150 plus 35 is a hundred and eighty-five and you've got the sum so this is kind of like a compact condensed way of writing a series let's look at number two so same idea here you're going to start at the bottom index your K in this case and you're going to put two in for K 1/2 to the second power is 1/4 times 2 is 1/2 say for example we put the next one in three we get 1/2 cubed which is 1/8 times 2 which is 1/4 say we put in 4 we get 1/2 to the fourth powers 1/16 times 2 is 18.8 until you get to the seventh term okay so here what you can see is that we're multiplying by 1/2 each time and we can see that the first term is 1/2 and now all we have to do is figure out how many terms are there from 2 to 7 now a lot of people mistakenly will look at this top number and say oh you know Mario it's 7 that's not quite right because we didn't start at 1 we started at 2 and another mistake students make is they say well oh seven minus 2 is 5 but there's actually 6 terms here because think about it if somebody says how many numbers are there from 1 to 5 5 minus 1 is 4 but really you can see there's 5 terms when you're counting the first and the last when you subtract you have to add one more so 7 minus 2 is 5 plus 1 more is 6 we have 6 terms we know it's geometric is our multiplying by 1/2 so we're gonna use this formula right here so the sum of these six terms the first term we said was 1/2 the ratio which is 1/2 there were six terms all divided by one minus one half so let's go to the calculator on this one and let's see what we get so we have one minus okay all divided by and I'm getting let's see if I can convert that to a fraction here for us it's 63 64 okay let's look at number three now this one you can see that it's an infinite okay some here we have a sum of an infinite number of terms here if we put three in we have one-third cubed which is 127 times four which is four twenty sevens if we put in four we're gonna get 1 over 81 times four which is 4 over 81 and dot dot now what's interesting about this one is it's an infinite okay number of terms we're going to use this infinite geometric sum formula but we want to make sure that that ratio is in between negative one and one right well you can see that this looks very much like our explicit formula here for geometric sequences a 1 times R to the n minus 1 you can see that that ratio is one third and you can test it out by writing a few terms but now if we want to find that some we say okay for over twenty-seven divided by a 1 minus the ratio which is one-third so we get 427 s over one minus one-third which is two-thirds of course when you divide by a fresh and it's like multiplying by the reciprocal we can do a little bit of cross reducing here three goes in here once 3 goes in here nine times so we're getting a sum of two ninths so if you like my teaching style and you want to learn more about sequences in series I'm gonna put a play out list right there going over more of these types of problems I'll see you over in those videos