 Well, welcome to episode 32. In this episode, we're going to continue talking about sequences and series like we did in the previous episode. However, today we're going to talk about two special sequence and series, those that are called arithmetic sequences and arithmetic series and geometric sequences and series. Let's go to our list of objectives and we'll see what we're going to be doing today. In just a moment, I'll be introducing arithmetic sequences. And then we'll look at two formulas for adding up arithmetic series. And we'll look at several applications of that. Then we'll switch over in the second half of this episode to an introduction to geometric sequences. And there are two formulas that are associated with geometric series. And again, we'll look at some applications of those. OK, now to begin with, let's look at what I mean by an arithmetic sequence. You know, if you're counting like 1, 2, 3, 4, 5, et cetera, you're just listing the numbers and putting commas in between. That would be an example of a sequence. And in particular, that's an example of an arithmetic sequence. Because you notice these numbers are increasing by the same amount. Namely, they're increasing by 1. So when I say 1, 2, 3, 4, my numbers are increasing by 1. So that's the characteristic of an arithmetic sequence. On the other hand, if I were adding up these numbers, like 1 plus 2 plus 3 plus 4, that would be a series, not a sequence, and that would be called an arithmetic sequence. Because again, the numbers are increasing by the same amount. Let's go to the first graphic or the next graphic and we'll look at a formal definition of an arithmetic sequence. A sequence is arithmetic if each pair of consecutive terms differs by the same amount. So I'm going to call the difference between consecutive terms d. So the first term I'll call a. The next term I'll call a plus d. So it differs from the first term by d. And then if I add on another d, I get a plus 2d. It differs from the previous term by d. And if I go to the nth term, I've added on n minus 1d. So right below this, this is a1. This is a2, namely a plus d. This is a3. And this term is a sub n. So you notice that in the second term, I've added on 1d. In the third term, I've added on 2d's. You notice the number of d's is always one less than the position number, because there wasn't a d in the very first term. So in the nth term, I've added on n minus 1d's. And I'm using the letter d here for difference in that case. So if I write a, I'll be referring to a1. And then, of course, d is the difference. Let's look at some examples of arithmetic sequences in the next graphic. Which of the following are arithmetic sequences? And if it's arithmetic, we're asked to find a and d in the nth term. Well, let's see. In the first example, we have 5, 8, 11, 14, 17. You notice this is a sequence, because I'm just listing the numbers. It's not a series. If it were a series, I'd be adding these numbers together. What appears to be the common difference of the numbers that we've written? Looks like it's 3. So it looks like a and d can be found in this sequence. So it looks like it's arithmetic. And I would say that a is, well, what would you say is a? a would be the first term, so that would be 5. And Stephen said the common difference appears to be 3. And now, what is the nth term? Well, if you remember, the formula that we just saw on the previous graphic said that the nth term is a plus n minus 1 times d. So in this case, that would be 5 plus n minus 1 times 3. Let me just clarify that a little bit so you can see n minus 1 there. Now, if you want, you could leave your answer in this form, or if you wish, you could multiply it out. If I multiply that out, that's 5 plus 3n minus 3. Or when I combine terms, that'll be 3n plus 2. It looks a lot simpler in this form, but I think in this form, we recognize the nth term of the arithmetic sequence. Now, for example, if I substituted in a 1 right there, how much would I get for a sub 1 if I plugged a 1 in right there? Wouldn't that be 0? So how much would this be? 5. It'd be 5. Yeah, I'm sure enough, that is a sub 1. If I substituted in a 2 right here, that'd be 5 plus 1 times 3 would be 8. Yeah, and that's the second term. And et cetera. So this would actually give me any of the terms in the sequence, assuming that this pattern continues. Now, if you look at the formula that I've reduced it to, 3n plus 2, look what happens if I substitute in a 1 in that expression. What will I get? 5. We get 5 again. So again, we get 5 for the first term. And if I substituted in a 2 in this expression, I'll get 8. And I get the second term. So you see, this formula certainly will give me the nth term. And this formula gives me the nth term. So you could write your answer in either of these forms. The latter one is reduced. But I think there's some advantage in leaving the answer in this form, just so that we recognize our general formula. OK, so the first sequence there is an arithmetic sequence. Let's look at the second one in part b. OK, in this case, let's say we start off with 1 tenth, and then we go to 8 hundredths, and then 600ths, and then 400ths. It looks like these numbers are getting smaller. And are they being reduced by the same amount every time? I think so, because you see, if I put a little 0 in here right after that term, then this is 0.10, 0.8, 0.6, or rather, 0.08, 0.06, 0.04. So David, what do you think d is in that sequence? 0.02. 0.02, and it'll be a negative, though, because we're reducing it. Yeah, so what we're doing is we're adding on negative 0.02. And a, the first term, is 0.1. I'll take that 0 back out over here. So it looks like this is an arithmetic sequence if this pattern were to continue. And the nth term, let's see, the term for a sub n would be a plus n minus 1 times d. And that would be 1 tenth plus n minus 1 times negative 0.02. So we're using decimals here. And also, we have a negative value for d, but it's still arithmetic. If I were to multiply that out, I would get 1 tenth minus 0.02 n plus 0.02. If I add those together, let's see, how much is 1 tenth plus 2 hundredths? 1,200ths? 1,200ths, yeah, exactly. Yes, that'll be 0.12 minus 0.02 n. So this is my reduced formula, but somehow this one doesn't look quite as pretty as that last one did in the first example. So you could give your answer this way, or you could give your answer this way for the nth term. OK, so the first two are both arithmetic sequences. Let's go to the third case, or the third example. In this case, we have polynomials. We have a variable included. So we have x, 3x, 5x, 7x. Does it appear that we're adding on the same amount every time? Yeah. Yeah, how much are we adding on? 2x. 2x. So that would make this arithmetic if that's the pattern that continues. And it looks like, in this case, a would be x, d would be 2x, because that's what we're adding on every time. And a sub n, let's see now, a sub n has the formula a plus n minus 1d. You remember the reason this is an n minus 1d is because in the nth term, I've only added on n minus 1d. So it wasn't a d in the first term. So the number of d's that have been added is always 1 less than the position number. And therefore, a sub n would be x plus n minus 1 times 2x. And if I multiply that out, that would be x plus 2nx minus 2x. And that would be, let's see, we need to remove the graphic, I think, up there. There we go. OK. And that would be 2nx minus x. And the last way I'll write this is to factor out the x and put it over here on the right, 2n minus 1 times x. Now in this form, I think we can certainly see the connection between this formula and this sequence. Because you see, if n is an integer, then 2n would be an even integer. So this would be even. 2n minus 1 would be odd. This is an even number minus 1. So this is an odd multiple of x. And that's what I have here, or odd multiples of x. So this certainly seems like a sort of a nice expression to reduce it to. Whereas up here, this seems quite a bit more complicated without the things reduced. But again, this shows us that it's in the standard form for the nth term of an arithmetic sequence. OK. Now let's look at some applications of arithmetic sequences, or at least some problems that pertain to them. And we'll begin with an example that we have on the next graphic. This problem says, suppose the second term of an arithmetic sequence is 14, and the fifth term is 50. What's the seventh term? OK, so the question's about the seventh term. So let's just say that these are the positions of the seven numbers that we have so far up to the seventh term. And the second term of the arithmetic sequence is 14. So I'll put a 14 there. And the fifth term is 50. And we want to know what is the seventh term over here. Well, let's see now. The first term is called A. And after that, I would say that this one is A2, and this one is A5. And the one over here is A sub 7. That's what we're looking for. And I know that A2 has a formula. A2, or in fact, A sub n, is A plus n minus 1 times d. Well, here's n minus 1. If n is 2, n minus 1 is 1. So that's just A plus d. I'll just take out the 1 right there. And I know that this is equal to 14. Now what about A sub 5? A sub 5 would be A plus how many d's? 4d. 4d. Yeah, that would be 4d, because I have one less d than I had the position. This is in the fifth position. And that's equal to 50. So you see, what I have here are two equations and two unknowns, two linear equations. We have A, we have d, and we have two linear equations that give their values. So why don't we subtract these two and solve for d? Now, of course, we could use matrix methods. We could use Kramer's rule, but for a 2 by 2, I mean, we might as well just use the methods that we've seen earlier. So when I subtract the A's cancel, and when I subtract here, I get a negative 3d, and when I subtract here, I get a negative 36. So that says that d is equal to what? 12. Yeah, d's equal to 12, exactly. Now, if I plug d in back up here, then that says A plus 12 is 14, and so A is equal to 2. So now I know A, so I know where the series begins, and I know d, these are increases by 12, and we want to find A sub 7. Well, let's just record this information over here on the side, and let's go back and calculate A sub 7. Now, you might say, Dennis, why don't you just fill in all the blanks, starting off here with 2 and add 12 and add 12 and add 12, and eventually you get over here to A to the seventh. What I'm thinking is what if this had been A to the 100th power, or A sub 100, then it would be difficult to write out all those terms. Let's see if we can calculate A sub 7 directly, and I'm thinking A sub 7 is equal to A plus what? A plus how many d's? 60. Yeah, do you agree with that? Everybody else agree with that? 60's? A plus 6d. And that's gonna be 2 plus 6 times 12. Yeah, what's 6 times 12? David, what's 6 times 12? Oh, 72. 72, so it's 2 plus 72 is 74. So we've decided that the seventh term is 74. Now, you know, we can check that answer by putting in a 2 here, add 12, add 12 again, add 12 again, add 12 again, we're at 50, we're at 50, what'd be the next number? 62. 62, and then add 12 again and get 74. And if you were to find the seventh term in this fashion, that would certainly be okay, but if I were to ask you for a term further out the sequence like A sub 20 or 30 or 40, this might be sort of inconvenient, and this is I think the quickest way then to find out that term. Okay, you have problems like this in the homework now where you'll first of all be given some sequences and you'll be asked, are these arithmetic sequences? Then you'll be given some information about a sequence like in this case, you were given the second term and the fifth term, and you had to predict a term further out. So that's fairly typical of your homework. Okay, now let's switch over to series. If I had an arithmetic sequence and then I added up the individual terms out to say the first in terms, if I added up the first in terms, that would be an example of an arithmetic series. And we have two formulas for adding up arithmetic series. Let's look at the next graphic and we'll see these. You'll find these in your textbook as well. If S sub n is the sum of the first in terms of an arithmetic sequence, now you see, here's the first term, here's the second term, third term, this is the nth term, but the difference now is we're adding these up, that makes it a series. So I'm gonna call S sub n the sum of the first in terms. Then there are two formulas that'll calculate S sub n. One formula says S sub n can be found with this formula n over two times two a plus n minus one times d. But then the other formula, which is a little bit shorter, says S sub n is n times a plus a sub n over two. Now that's the first term plus the last term added together and divided by two. Let's first see where these two formulas come from and then we'll look at how we can decide which of the two formulas we want to apply to certain applications. Okay, if you're coming back to the green screen, let's just look at this problem of the question of where those formulas have come from. So we said that we were going to define S sub n to be a plus a plus d plus, let's see, the next term would be a plus two d. And let's just keep going until I get to the nth term and the nth term is a plus n minus one times d. Just like we were using a moment ago when we were calculating the nth term of a sequence. Okay, now what I'm gonna do next is I'm gonna write that summation down backwards. I'm gonna put a over here and then just before it I'll put a plus d and then if I reverse this order, I'll have the nth term up here at the front. I better move this back just a little bit to give me some more room to fill that in here. That's gonna be a plus n minus one times d and I have all these other terms written in here. Now, if I add on both sides, if I add on the left and if I add on the right, then on the left I have S sub n plus S sub n equals two S sub n. And if I add here, I have a plus a plus n minus one d. That's gonna be a total of two a plus n minus one d. And if I were to add here, you notice this time I have an extra d in the first term but I have one less d in the term just below it. So when I add those together, I'm gonna have the same number of d's that I had over here because I've increased the d's there but I've decreased the d's here because these d's are going down. So when I add these together, again I'm gonna get two a plus n minus one times d. Now let's go to the very last summation down here on the very end. When I add these two together, what will I get? Susan, what will I get there when I add those together? No idea. A plus n minus one d plus a, how much is that sum? Two a plus n minus one d. Exactly, yeah, two a plus n minus one times d. Okay, well the idea is I have exactly the same sum in every one of these columns and this column and the next column and the next column, these all look alike. And the reason is every time I add an extra d, I've deleted a d on the other row. So when I put them together, I still have the same number of d's. So this tells me I have a multiple of two a plus n minus one d's. Now I have as many of these expressions as I had terms listed up here. And remember that was the sum of the first n terms. So I have n times two a plus n minus one d. I have n of them added together, so I'll just multiply by n. So this tells me that s sub n is equal to n over two times two a plus n minus one times d. And this is the first formula that we saw in that previous graphic. If you could, let's get ready to go back to that graphic in just a moment. But there was another formula that was given up there. Let me show you where that one comes from. If I just continue this, I'm gonna separate the two a and write it this way. n over two times a plus a plus n minus one times d. Now where have you seen this before? What is that? That's the last term of the series. Right, that was the last term of the series. This is a sub n. So what we have now is n over two times a plus a sub n. Okay, now the two formulas that we showed in our graphic just a moment ago were the formula given right here, n over two times two a plus n minus one d. And the other formula was n over two times the sum of the first term and the last term. Now, these are both helpful formulas in solving problems, but we can't use them interchangeably all the time. You see, if I don't know the last term, then I couldn't use this formula. But if I were given an arithmetic sequence, I should know the first term, so I know a, and I should know what d is. And therefore, and if I knew how many terms there were in, then I'd put in minus one here. So sometimes I'll want to use this formula, but if I know the first and the last terms, I might want to use that formula. Now let's go back to that last graphic and just look at those two formulas one more time before we go to some examples. Yeah, you see it says a sub n was n over two times two a plus n minus one d. You'll find that formula in your textbook. And then the second formula was s sub n equals n over two times a plus a sub n. Yeah, so okay, now let's go to the next example and we'll see how to apply that. Use one of these two summation formulas to find the sum of four k minus one where k goes from one to 20. Okay, now let me just write that again right below here. We're taking the summation k equals one to 20. Of four k minus one. Now the first question we might ask is this an arithmetic sequence or rather series? Because if it's not arithmetic, I wouldn't be able to use one of those formulas. Well, what would the first term of this summation be when you substitute in a one? Three. It'd be three, yeah, if I plug in a one, four minus one is three. And when I substitute k equals two, David, what would you get for k equals two? Okay, equals two, you'd get a k. Well, if k is two, that'd be seven, yeah. Right. And if I substitute k equals three, Mike, what would I get when k equals three? I'm still working on it. Okay, let's see, four times three minus one, that'd be 11. Yeah, and it looks like what's happening is these numbers are increasing by four. And you notice there's a four coefficient right here. And if I go to the very last term, when I substitute in a 20, that'll be four times 20 minus one, 80 minus one is 70, 79. Okay, so the question is what is this sum? Well, according to my formula, I would have called this S sub 20, the sum of the first 20 terms. And if I were to use the second formula, N over two times A plus A sub N, that would be 20 over two because N is 20. And the sum of the first and last terms, that would be three plus 79, three plus 79. And that's gonna be 10 times 82, which is 820. That's the sum of all these numbers. So without writing them all out and then physically adding them up, I've used a formula to compute the sum to be 820. Now, what if I were to solve this using the other formula? I think we'll still get 820, let's just try that. So in this case, we have S sub 20 and the formula is N over two times two A plus N minus one times D. And how much do we say N was? 20. 20, yeah, that'll be 20 over two times two A. Well, let's see now, A is three, so two A is six. And then N minus one, well, we said N was 20, so that'll be 19 times D. David, what's the value of D in this problem? What's the common difference? What's the common difference up here? Four. It's four, yeah. So we now have 20 over two is 10 and six plus, now four times 19 is, let's see, I think that's 76. So this is 10 times 82 once again and that's 820 just like before. So we get the same answer using either formula in that case. Okay, one more example and then we'll move over to geometric sequences and series. Let's go to the theater problem. It says a movie theater auditorium with stadium seating has 25 rows of seats. If there are 22 seats in the first row and 46 seats in the last row, what is the seating capacity of the auditorium? Now, it doesn't mention this, but this is intended to be an arithmetic sequence. Every row is increasing by the same number of seats. So let's say that here's the first row, then here's the second row, then here's the third row, and then here's the fourth row, et cetera. And we keep on going until we get to the 25th row. So this is row number 25 and this is row number one up here. And on the first row, there are 22 seats. And on the last row, there are 46 seats. That's a difference of 24. So what would you guess is the increment of seats on every row? What does it seem to be the increment in seats here? Each row is increasing by the same number of seats. How much would you say? There are, it's a 24 difference from 22 to 46. And there are 25 rows. I'm thinking that G is probably gonna be, what do you think? One. Is one, yeah, I think D is one. And you might say, but 24, shouldn't it be 25? We'll see, there is no D added on the first row. The Ds only begin on the second row. That's why this number is one less. So if I use a formula, I wanna find S sub 25, the sum of the seats on 25 rows. And we have the first term and the last term, so I think the simplest formula to use here is N over two times A plus A sub 25. And while we're at it, why don't I put a 25 in where the D is? Because that's the value of N, 25. So this is gonna give me 25 over two times, what would be A in this case? 22. Be 22, yeah, that's the number of seats in the first row. And then we're gonna add on the number of seats in the last row, that's 46. And so this is 25 over two times 68, which is 25 times 34, 25 times 34. Now let me ask you, if you had 34 quarters, how much money would you have? 34 quarters, 34 quarters. Well, let's see, if you had 32 quarters, that would be eight dollars. And you'd have two quarters left over. So this would be 800 and two quarters left over 850. So I think there are 850 seats in this theater. Now this problem could have been worked using the other formula too, but in that case you have to use the value of D rather than the value of the last term. Okay, in your homework, you will see problems then on arithmetic sequences and series. And the sort of formulas you need to keep in mind are the formula for the nth term of the sequence and the two summation formulas for the arithmetic series. Okay, we're gonna shift gears now and look at another type of sequence, and this is a geometric sequence. And the distinction between geometric and arithmetic is whereas in arithmetic sequences, the numbers in the sequence increase by the same amount, they have a common difference. In this case, they're gonna increase by a common multiple. Every consecutive term is the same multiple of the one just before. Or you may say the ratio of two consecutive terms is always the same. Let's go to the next graphic and we'll see this. Here's an example of what I call a geometric sequence. A sequence is geometric if the ratio of consecutive terms, a sub n plus one over a sub n is constant. Okay, so the first term again I'll call a, and you notice now we multiply by an r. I'm not adding on D, but I'm multiplying by r, so I get ar, then I multiply by r again and I multiply by r again until I get out to the nth term. Now, if I take the ratio of these two terms, ar over a, you notice that ratio reduces to r. And if I take a ratio of these two terms, I have ar squared over ar, so I get r again. So consecutive terms have a common ratio, and that's why we're using the letter r for the multiplier because it represents the common ratio of that sequence. Okay, and this increases, or this series changes, I should say, geometrically. Now, let's go to the next graphic and we'll see some examples of what are geometric sequences and there may be one in here that isn't. Okay, this example says which of the following are geometric sequences and if it's geometric, find a, find r, and find the nth term. Well, this time what I'm looking for is not a common difference, but a common ratio. So what is the ratio of the first two terms? Mike, what's the? It's five. Yeah, Mike, what's the ratio of the second two terms? And that's also five. And I think the last one's five. So it looks like a is two and the common ratio is five and that becomes the multiplier. Two times five is 10. 10 times five is 50. 50 times five is 250, et cetera. So it looks like in our first example here, a is equal to two and r is equal to five. Now, when it comes to the nth term, a sub n, well, that'll be a times, and then I've multiplied by a bunch of r's, but you know, there's gonna be n minus one r's that I've multiplied by and that's because there was no r in the first term. The r's were introduced in the second, the third, the fourth term. So I always have one less r than the position number. So using that formula, this'll be two times five to the n minus one power. This is a formula for the nth term. Now, if you wanted to know, for example, what is the 10th term of this geometric sequence? If you wanted to know a sub 10, you'd put a 10 in here and that would be the ninth power. Two times five to the ninth power. It'd be a very big number. You'd probably want to use a calculator to find it, but that's how you would find the 10th term or any other specific term of that sequence. Now, looking at the sequence in B, let's keep that, yeah, there we go. Okay, we start off with one and then we have the square root of two. What's that all about? And then we have a two and then we have a two squared to two. Is there a common ratio? Well, let's see, what's the ratio of these two terms that that would be the square root of two over one is the square root of two. What's the ratio of these two terms? That would be two over the square root of two and can I reduce that? How would you reduce that? Multiply both the top and bottom by root two. Yeah, exactly. You remember if you have a radical in the denominator, the way you reduce it is you multiply by another square root of two on top and you multiply by a square root of two on bottom. And what's the square root of two times the square root of two? Two. Two and those two's cancel and this is the square root of two. Well, things are looking hopeful here. I have the same ratio so far. What's the ratio of these two terms? Square root of two. Square root of two again. So if the pattern that we're seeing is the pattern that's actually intended to produce the next terms, then this is a geometric sequence as well. If you think back to the last episode, you may remember there was an example where it started off one, two, three and I said what was the next number and it seemed like the obvious number was gonna be four but I showed you some formulas that could actually make the next number be anything you wanted. So we can't guarantee that this pattern is gonna continue to be geometric but if we assume it is, it certainly looks like it's geometric from all the information that we're given. Okay, now the questions we were asked here were to find A, find R and find A sub n. Let's see, Susan, what would be A in this case? One. It would be one, okay. And what would be R? Square root of two. Square root of two, exactly. And now who can tell us what is the nth term? The formula is A, R to the n minus one. So that's gonna be? One times the square root of two to the n minus one power. To the n minus one power, okay. And you know, there's really no reason to show the one in front. So we could say just the square root of two to the n minus one power and if we wanna be fancy we could say this is two to the, let's see, that's two to the one half power. And one half times n minus one is n minus one over two power. You could write it that way. This answer is fine right there. This answer's fine and this answer's fine. So you don't have to reduce the exponent but you can see this does look simpler than any other. And I've just changed the square root into a one half power and then multiplied by exponents there. Okay, let's look at the third sequence. We have the third sequence in that same graphic. There you go. Okay, we have one half, one sixth, one 24th, one over 120. Let's see, what's the ratio here? One sixth over one half, one sixth over one half. Well, if I invert and multiply that's one sixth times two over one is a third. That's the ratio of the first two terms. What's the ratio of the next two terms? One over 24. Divided by one over six. Yeah, Mike's already reduced it. What'd you get? A fourth? One over four. Okay, so immediately, Mike, what does this tell you about this sequence? It looks like the next one's gonna be a fifth. Next one's gonna be a fifth but I guess the question is, is this geometric? Is that a geometric sequence? No. No, it's not, because we don't get the same ratio. We're getting a pattern in the ratios, one third, one fourth and I think the next one is one fifth but it's not the same ratio. So it's not geometric and therefore we don't have to find a, we don't have to find r. There's no nth term for us to find but only if it were geometric. Okay, let's go to the example, or the next example. If the first term of a geometric sequence is six and the fourth term is 81 over four, what's the fifth term? This sort of sounds like the problem we were asked about arithmetic sequences a minute ago. We're given two terms and we're asked to find some other term. So let's see, there are five terms mentioned at all so I'm gonna put down five blanks here. The first term is six. Well, that tells us a right away. The fourth term is 81 over four and we wanna find out what is the fifth term. What is the fifth term? Well, let's see, we have a is equal to six. We have a sub four is equal to 81 over four. Now, you know there's a formula for the fourth term and that would be a times r to the, what power should I put on that r for the fourth term? Three. Third power, yeah. You remember for a sub n, I put n minus one r's. So for a sub four, I put three r's. So that's gonna be six r cubed. Okay, well, let's pull that out and solve it. We have six r cubed equals 81 over four. That says that r cubed is equal to 81 over six times four. Now, you know, we could cancel a three there and that's gonna say r cubed equals 27 over what will the denominator be? If I cancel off a three. Eight. That'll be an eight, yeah, that'll be an eight. And so r will be the cube root of that, which is three halves, r is three halves. Okay, we have a, we have r. So let's find a sub five. Well, you know, probably the quickest way to find a sub five is just multiply the fourth term by three halves. So the fourth term, let's 81 over four. And if I multiply it by three halves, I'm gonna get 243 over eight. So it looks like this is 243 over eight. But the other way to find that term, just in case I weren't asked the next consecutive term, where it wouldn't be quite so convenient to do that, to find the fifth term, I could solve it this way. A sub five is a r to the fourth power, which is six times three halves to the fourth power. Which is six times, let's see now, you know, three squared is nine, what's three to the fourth? If three, this is three squared times three squared, three to the fourth. 81? Yeah, yeah, what I'm thinking of is this is, when I say three to the fourth, that's three squared times three squared, that's nine times nine. So nine times nine is 81. So that numerator is 81. And two to the fourth, well, that's two squared times two squared. How much will that be? 16? 16, right, very good. Now, if I cancel off a two, I have three times 81 over eight. And that's gonna give me two, 43 over eight. And that's exactly what we got up here. Now, clearly this is the longer way to get to the answer. If you know the fourth term and you wanna get the fifth term, just multiply by r. But if you had been asked to find some term further down the line, this might be quicker than finding out all the terms in between. Okay, so the question was, what's the fifth term and it's 243 over eight? Okay, let's go to the next example. Oh, actually it's two formulas. Now, we come to the question of how do you add up geometric series? Well, the two formulas that we wanna use, first one says, if you add up the first n terms of a geometric series, that would be a plus a times r plus a times r squared up to, here's the nth term, which can be abbreviated this way, then that summation will be a times one minus r to the n over one minus r. Now, we wanna verify that formula here in just a minute, but this is how you add up the first n terms of a geometric series. You know, the summation here goes from one to n, and this is a, r to the k minus one power, because in the kth term, you have one less r than you have the position. That's how all of these have been going. Now, there's a separate formula when you have an infinite series. If the absolute value of r is less than one, that is if r is somewhere between negative one and one, and you wanna add up an infinite geometric series, so you notice this just keeps on going, there's a dot, dot, dot over here. This goes from one to infinity, so k goes from one to infinity, a, k to the k minus, a, r to the k minus one power. Then we have this formula for the summation of the series, a over one minus r. Now, let's just see where those two formulas come from, and then we'll apply them to some applications. Okay, so we're talking about adding up, first of all, the first n terms of a geometric series, that's a plus a, r, plus a, r squared, and when we come out to the nth term, we said the nth term was a, r to the n minus one, but there's a formula that we were given on that previous graphic, let's see where the formula comes from. What I'm gonna do is multiply both sides here by r, multiply by another r, and so I'll multiply each of these terms by r. I'll get a, r, after multiplying this guy by r, and then for the next one, I'll get a, r squared, and if I keep going, you notice what happens is all these terms, all these terms just get pushed over by one and their exponent goes up one, so I'll have an a, r to the n minus one here, plus a, r to the n, that's where I multiply the very last term by r, so I get a, r to the n. Now, if I subtract these two, if I subtract these two, I have s sub n minus r, s sub n, and when I subtract over here, I get a lot of cancellation, these cancel, these cancel, everything in the middle cancels, so what I have left is a minus a, r to the n. If I factor out the s sub n, in parentheses, I have one minus r, and over here, if I factor out an a, I have one minus r to the n. So this tells me that s sub n is a times one minus r to the n over one minus r. So if I know the initial term of a geometric, of a finite geometric series, and if I know the number of terms in the series, then I can calculate the sum of the terms in the series by using this formula right here. Let's just take an example of this before we look at the other formula. Let me just write that formula here at the top of the green screen, and I'll show you an example of how this would work. Suppose I were gonna add up one plus two plus four plus eight plus 16. Now I think this is a geometric series. What would you say is a in that series? One. One, and what's the multiplier in each series? Two. Two, yeah, or you could say, what's the common ratio of consecutive numbers? Two over one, four over two, eight over four. It looks like r is two. So what is the sum of that series? Well, that sum would be the sum of the first five terms, s of five. And we said we have a formula for that, and it looked like a times one minus r to the n over one minus r. Well, a is one, r is two, and n is five, so that's one plus one minus two to the fifth over one minus two. And that'll be one minus 32 over negative one, which is minus 31 over minus one, which is 31. 31 should be the sum of these numbers right here. We get 31. Now, of course, we're using a formula to get this answer. Let's try adding them up a different way. Look, here's a 10 right there, and here's a 20 right there, and then there's one more. So I've got 10 and 20 and one, sure enough, that makes 31. Now, you might say, well, Dennis, you've just managed to make a mountain out of a mole hill. It was really easy to add those up. Why did you go through all this trouble to find the sum of those numbers? Well, of course, if there are only five numbers in the series, then it's not difficult to add those, but if there were many more numbers in the series, this formula would allow me to calculate that sum anyway. Okay, can we go back to that last graphic with the summation formulas on it? I just want to look at the other formula that's on there. Here we go. You remember we said that if I'm adding up an infinite, if I'm adding up an infinite geometric series, that the formula is s equals a over one minus r. Okay, now let's just see where that one comes from. If you come back to the green screen, this said, we said that s sub n was a times one minus r to the n over one minus r. Now, let's assume that r is between negative one and one. Or the way it was written in the graphic, the absolute value of r is less than one. Now, in other words, this number might be a half or a third or negative a fourth. What happens when you take a number between negative one and one and raise it to larger and larger and larger powers? What's going to happen to that? You get smaller and smaller. Yeah, this is going to approach zero. So when I look at s sub, maybe I should say infinity or on the graphic, I think I just called it s, as the number of terms becomes infinite, this nth power goes to zero, and so I have a times one minus zero over one minus r, which is a over one minus r. That's the formula for adding up an infinite geometric series, but you can only apply it in the case where r is between negative one and one. Okay, let's go to the examples and look at how we could apply those last two formulas. In the first example, it says we're going to find the sum of one minus a half plus a fourth minus an eighth plus, it's a dot, dot, dot, and this is minus one half to the n minus one power. You see, it looks like every time we're multiplying by negative a half, and that causes the signs to alternate. So I put a negative in my multiplier right here. A is one, so you don't see a in front. R is negative a half, and we raise it to the n minus one power. Now what's the sum of this series gonna be? Well, we have a formula for this, and that is that we're supposed to put a times one minus r to the nth power over one minus r. Well, a is one, we've got that over here, a is one, so I can forget that term. R is negative a half, and I'll put negative a half right here. So if I just replace those, if you come to the green screen, I'll have room to write that out. If I replace those with the number of values, we said that a was gonna be one, r was gonna be negative one half, and we wanna calculate the sum of the first n terms, and that's a times one minus r to the n over one minus r. That's one minus negative a half, and I'll substitute for r right here. That was a negative one half to the nth power, and so a was one, so we have one minus negative one half to the nth power over three halves, one plus a half, and if I invert and multiply, this is two times one minus negative one half to the nth power all over three, and that's the sum of that, that's a formula for the sum of that first finite series that we were given in that example. Okay, let's go back to that graphic and look at the second example. You notice in the second example here, we have an infinite series, because this just keeps on going. There's a different formula for this, and the formula is that we should use s equals a over one minus r for this infinite series, and I can use this formula if r has absolute value less than one. Well, let's see, what is a in this problem? What is a? One fifth? No, not a, a is the first term. So a would be two, yeah. Now, what are we multiplying by every time to get the next term? One fifth. Yeah, that's where the one fifth comes in. So one fifth has absolute value less than one, and therefore the sum of this series is gonna be a over one minus one fifth. Now, I believe if you reduce that, I don't have much room here, but I think you're gonna get five halves. That's the sum of this infinite series. Okay, now, let's look at this last example, and this isn't actually a series, this is a decimal number. Now, you're probably familiar with a decimal, if you put a bar over digits, what that means is those digits keep repeating, so this is 0.81, 81, 81, 81, because the 81 keeps repeating. Let's come to the green screen, and I'll show you how we'll convert this to an infinite series and figure out what rational number this is. We have 0.81 with a bar over it. Now, that means 0.81, 81, 81, and that goes on forever. Now, I can write this as a series if I say 0.81 plus 0.0081 plus 0.00081, and you see what that does is it just keeps shifting the 81s over a little bit. It looks like a is gonna be 0.81, and it looks like the multiplier here is 100th. Every time I'm just moving the decimal back two places or forward two places, so that means I'm just multiplying by a power of 10, and this is an infinite series, and so 0.81 with a bar over it must be, that's the sum, must be a over one minus r, which is 0.81 over one minus 0.01, which is 0.81 over 0.99. Okay, so now if we just reduce that, we'll know what is the rational number that is this decimal. That's 81 over 99, and if I reduce that, if I divide by nine, I get nine over 11, nine over 11. So this repeating decimal is the same thing as the fraction nine over 11, and I found it by separating this into individual pieces and noticing that every time I had a common multiplier to get to the next term. Okay, let's go to the next graphic, and this problem ought to get your attention, I think. Here's a way that you can become actually quite rich. It says, congratulations, you've been offered a job that pays one cent on the first day, two cents on the second day, four cents on the third day, et cetera, so that your pay continues to double every day thereafter. What is the total amount you would be paid for 30 days? Now, let's go to an interview that I had with some students out in front of the library where I asked them a question very much like this. Well, let's see, class, we're just outside the library, and I have a really interesting question for you here. Which one would you choose? So you'd be thinking about this. So let me ask a student here what she would do. Tell us your name. My name is Yurango. Yurango, and what is your major, Yurango? Drafting technology. Drafting technology. Well, you could possibly be very rich with this plan. What would you choose, $100 a day for 30 days, or one cents, two cents, four cents? I think I'll choose second. You'd choose the second one, why did you choose that? It sounds so small. No, it's not small. Oh, it's not small, okay. On a fourth day, it will be like 16 cents. 16 cents and then 32, yeah, it starts to grow very fast. You know, actually, you've chosen the one that would pay you more money, but let me change it. What if I were to give you $1,000 a day every day for 30 days, let's see, for 30 days, it'd be $30,000. Would you still choose this, or would you rather go with a thousand? A thousand. You'll go with $1,000 a day? It's 30 days. 30 days, $30,000, that's a lot. Billy, what's your major? I'm majoring in biology. In biology, and I think you just told me a minute ago that you're in college algebra right now. I am taking college algebra. Yeah, so in other words, this is in his course, although I bet he hasn't gotten to it yet in his course because we're filming this a little bit earlier than the actual studio lecture. So Billy, here's the choice, $30 a day, whether $100 a day for 30 days, or one cent, two cents, four cents, et cetera. What would you choose? Well, I think definitely I would choose this one. Why is that? Well, it multiplies very fast. It, I don't know, expounds exponentially. It goes up exponentially, good words, okay? So by the time you get towards the end of the days, you're making a great deal more than $100 a day. Yeah, yeah, it starts off pretty low, but it does get to be bigger. So I'll tell you what, what if we changed it to $1,000 a day? $1,000 a day, it's pretty close. What, $30,000? Well, I think I would have to add up the totals and for 30 days, $30,000. And then- What's your gut instinct on that? I would probably just go with this because it multiplies a lot quicker. Yeah, yeah. And anytime that it goes to an exponent, it's gonna go a lot faster. Yeah, okay, Billy, here's my last offer. All right. What if I gave you $10,000 a day for 30 days? What's that, $300,000? Where can I get a job like that? Pay me $10,000 a day, sounds good. Would you rather do that? Just off my gut instinct, I'd probably do this just because you might be earning millions, who knows? As a matter of fact, Billy, this, you'll make about $10 million here. So hey, thanks a lot. All right. Very good. Thanks. Okay, let's see why that would be over $10 million. If we go to the green screen, let's say on the first day, you'll be paid one cent. So I'll write that as a decimal. On the second day, you'll be paid twice that much, two cents. And then four cents, et cetera. And we're gonna go out here until we get to the, this is the 30th day right here. I won't bother figuring out how much that is. But I wanna calculate the sum of this finite geometric series. You see, it is geometric because A is one cent and R is equal to two because we're doubling every time. And to find S sub 30, our formula is to say A times one minus R to the N over one minus R. And I'll be substituting in 30 for N. I'll put in two for R and I'll put in 0.01 here for A. So that'll be 0.01 times one minus two to the 30th power over one minus, one minus two. And that's gonna be 0.01 times. Now let's see, you know, this is gonna be a negative one right here, so why don't I just reverse the order up on top and say two to the 30th minus one. Now I'm just about out of time, but I tell you what, I'm gonna leave this to your devices, to your calculator, you and your calculator to find out that this is more than $10 million. You'll just have to raise two to the 30th power, subtract one, and then multiply by 0.01 because you were getting one cent on the very first day. Okay, I think we're out of time. I'll see you next time for episode 33.