 Hey, welcome to episode 31. Today, we're talking about a new set of topics. And we just finished talking about the conic sections, parabolas, ellipses, and hyperbolas. Now we move on to something new. And for the next two episodes, this one in the next episode, we'll talk about sequences and series. And then we'll look at some applications of sequences and series and episodes that follow that. Let's look at our list of objectives for this episode. First of all, we'll have an introduction to sequences to see what sequences are. And then we'll look at a special sequence called the Fibonacci sequence. It's a sequence that's about 800 years old. Then we'll have an introduction to series. We'll look at notation for abbreviating series called sigma notation. And then finally, we'll conclude with properties of series. OK, well, let's begin with sequences. OK, let's go to the green screen and look at an example of a sequence and see what we can discover about it. Suppose we have the sequence 2, 4, 6, 8, et cetera. Now, you know, the number in the first position is referred to as the first term. And then the number 4 in this case, that would be the second term and the third term, et cetera. Now, of course, that gets a little cumbersome to write that out. So there's another way of explaining this. The number that's the first term is usually abbreviated as a sub 1. There's nothing special about a. You could call it b sub 1 or whatever. But the subscript indicates this in the first position. The number in the second position would be a sub 2 and then a sub 3. And it looks like a sub 4 is equal to 8. It's a sub 4. Now, what would you guess is a sub 5 in the sequence up here? Yeah, it looks like that ought to be 10. Because somehow we're just doubling. We're just writing down the even numbers in this case. But what really fixes this sequence so that I know exactly what all the numbers are is to have a formula that will tell me how to generate any particular term. Now, if I were to go further out in the sequence I've written here out to the, let's see, that's the first term, the second term, the third term, what if I were to go out to the nth term? I'll call this the nth term right here. This is where n is some positive integer. Then this would be called a sub n. If I have a formula for a sub n, then with that formula I can predict any term of the sequence that I want to predict without actually listing all the numbers out to that position. For example, you notice that this number is 2 times 1. This number is 2 times 2. This number is 2 times 3. And 3 is the position number. This number is 2 times 4. And it's in the fourth position. 2 times 5. So if I were to go out to the nth term right here, I would say that the nth term is 2 times n. 2 times n. Now you see if n is 1, 2 times 1 is 2. And if n is 2, for the second position, 2 times 2 is 4, et cetera. And so for the sequence that I've written as our example, I can summarize that sequence by merely giving a formula for the nth term. And I can say that a sub n is 2 to the n. Now you see with this formula, I could ask you now, what is a sub 5? What is a sub 5? That is the number that goes in the fifth position. And that would be 2 times 5 or 10. And that's exactly what we saw right there. Can anyone tell me what is a sub 1,000? 2,000. It's 2,000. Right. See now we can just predict these terms very easily once I know what the nth term is. Let's go to the next graphic. And we'll look at an example of how we use the nth term. The nth term definition. In this example, it says write the first four terms and the 100th term of the sequence whose nth term is given by each of the following. So first of all, what if a sub n were equal to n squared plus 1? And I want to write down the first four terms of that sequence. And then, what's the 100th term? Well, to find the first term, that means a sub 1. So I'm looking for a sub 1. So what I do is I plug in a 1 for n right here. And what do you get when you evaluate that? You get what? You get 2, yeah. And to find a sub 2, I'll just substitute a for a sub 2. I'll put in a 2 right here. And how much will that be? That'll be 5. And what will be the next term of the sequence? 10. 10. Yeah, that'll be 3 squared plus 1 is 10. And the fourth term is? 17. Yeah, 16 plus 1 is 17. So you see, the purpose of having the fifth term is it makes the sequence, shall we say, well-defined so that I can pick out any term and I can figure out what it is. Now, if I go further out, what is the 100th term out here? Well, let's see. Now, that would be, let's write it right below here. A sub 100 would be 100 squared plus 1. What is 100 squared? 10,000. 10,000. Yeah, you remember in the last episode we talked about squaring numbers that end in 0s. All you do is just square the lead number, square the 1, and then double the number of 0s. So I'll put down 4 0s plus 1. That's going to make 10,001. So this number is 10,001 that goes in the 100th position. OK, let's look at part b and see how we'd solve it. Now, this formula is a bit more complicated in part b. Let's go back to that graphic. In part b, I have a product of two factors and each one of them has an n in it. There's an n in the exponent and there's an n in the second factor and this is what we call the sequence b sub n. So the terms are b1, b2, b3, et cetera. And we'd like to find the first four terms and the 100th term of the sequence. What would be b sub 1? Yeah, let's see. If I put a 1 in there, negative 1 to the first power, that's a negative 1. And if I put a 1 in here, that's 2. So I get negative 2. Does everybody agree with that? What would be the second term? 3. 3. Let's see. If I put in a 2, let's see, for b sub 2, that'll be negative 1 squared. And then 2 plus 1 is 3, so that'll be positive 3. It's a positive because when we square the negative 1, that negative 1 essentially becomes insignificant. What is b sub 3? Negative 4. Negative 4, right, negative 4. And b sub 5 is 5. You notice how the signs of the sequence here alternate? Negative plus, negative plus. That's because this coefficient right here alternates between a negative 1 and a plus 1. As you keep raising the power by 1, you keep changing the sign. This term in a sequence is sometimes referred to as an alternator. You know on your car, you have an alternator? Well, a series can have an alternator. And what it does is it makes the sign alternate. So if I had left that portion out, this would have just been 2, 3, 4, 5. But the alternator forces the signs to change. By the way, what is the 100th term going to be? First of all, will it be plus or minus? It'll be plus. Yeah, because if I substitute in a 100, that'll be a plus 1. And so what number will this be then? 101. Exactly. OK, so here's an example of an alternating sequence. Now when I come to part c of this example, we don't have an alternator. So we'll just evaluate the nth term that we're given. And this is the c sequence. So we'll call these numbers c1, c2, c3. And we want to find the first four terms, first of all. Well, for c1, what number would we get? Matt, what would be c1 here? That would be 1. 1, 2. 1, 1, 1, 1, 2. Yeah, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1. And Linnae, what would be c2? 2 thirds. 2 thirds, yeah, 2 thirds. And the next number, anybody? 3 fourths. And then 4 fifths. Now you see what this sequence, the way the sequence is defined, is I have an integer over the next largest integer, n over n plus 1. So I have 1 over 2, 2 over 3, 3 over 4, 4 over 5. You notice that the numerators are 1, 2, 3, 4. Because the numerator here was just n. Whatever is the position number, that's what you put on top. The denominators are 2, 3, 4, 5. They're 1 larger, and they begin on a 2. And on top, I begin on a 1. So basically, I see two different sequences in the numerators and in the denominators that go together to make these fractions. And what is the 100th term? 100 over 101. 100 over 101, exactly. OK. Now, you might think that sequences are actually rather simple, but if we look at the next example, I think we'll see that there's actually some subtleties about sequences that might surprise you. In this graphic, it says, is the next term obvious? OK, now here we have a sequence, 1, 2, 3. And the question is, is the next term obvious? And you might say, well, of course it is. 1, 2, 3, 4, 5, like we just talked about a moment ago. Well, you know, actually, this depends on how you define the nth term. Now, if the nth term is defined as a sub n equals n, then you see a sub 1 is 1. Yeah, that's right. a sub 2 is 2. Yeah, that's right, a sub 2. a sub 3 is 3. What would be a sub 4? Would be 4. And a sub 5 is 5. So that's what we would be expecting. But suppose a sub n were defined this way. If I calculate a sub 1, I'd put in a 1 right here. I'd put in a 1 right there, and that's going to be 0. It doesn't really matter what the rest of those are. That's going to be 0. So when I add it up, a sub 1 is 1. What happens if we substitute in 2? What is a sub 2? Well, let's say I'll put a 2 here. And this term will be 0. So it doesn't really matter what the others are. When I add this up, I get 2. Yeah. And a sub 3 is 3 plus, there's a 0 down here, so we're going to get 3. But now things go sort of awry. What is a sub 4? 14. Yeah, if I plug in a 4 there, and then if I plug in a 4 here, that'll be 3 times 2 times 1. And we get how much? We get 10. A sub 4 is 10. And what would be a sub 5? Let's see if I put in a 5 here, plus, that'll be 4 times 3 times 2. 4 times 3 times 2. That's 24 plus 5 is 29. OK. So back up in this sequence, when I said what's the next term, you could have just as easily have said the next term should be 10, and then the next term should be 29 if, for some reason, you thought this was the pattern. So my point is this, you can actually make the next term be anything you want. And when you give a few terms of a sequence, the next term is not obvious. And what you have to know is a formula for the nth term before you can guarantee what the next term of the sequence will be. Now if we look at part c, it says if we want the fourth term to be some number k, whatever k might be, we can make this be k if we use this formula. a sub n plus k minus 6 plus this product. n minus 1, n minus 2, n minus 3. Now you might say, well, why is it k minus 6? Well you see, if I were to calculate a sub 4, a sub 4, if I plug in a 4 here, I get 3, 2, and 1. What is 3 plus 2 plus, 3 times 2 times 1? 6. So if I take k minus 6 and then add on that 6, this answer will end up being k for the case a sub 4. So you can actually create rather complex formulas that'll make the next term be whatever you'd like for it to be. You might say, Dennis, can you make up a formula that you can make the next term here be, say, 100, and the next term after that be negative 10? Yes, there are ways that you can design these formulas that can actually produce several predictable, or should we say unpredictable, numbers in that sequence. OK, so I just wanted to point out that these sequences are not necessarily as obvious as you might think and that you can actually create formulas that will surprise people as to what the next number would be. OK, let's go to the next graphic. OK, now after just seeing that when you're given a sequence, you can't always predict what the next number is going to be, let's just give it a try to see what would be the obvious next number in a sequence. So here's an example in which it says, for each of the following sequences, find the next term and find a formula for the nth term. Now, you know, from what we just said, this next term could actually be anything we want, but I think there's a fairly obvious sequence here that we could predict. So find the next term of the sequence. Well, class, what do you think that should be? 11. 11 is the obvious choice. OK, now can we find a formula for the nth term? Well, you know, although it seems to be fairly clear what this sequence is intended to be, coming up with a formula for the nth term may not be quite so obvious, so let me show you how you could come up with a formula for this. I'm going to make a graph. And I think I'll only draw the first quadrant. And here's one, here's two, here's three, here's four. So this is my choice for n. And then on the vertical axis, I'm going to plot the value of a sub n. So in other words, when n is one, a sub n is also one. So when n is one, a sub n is one. I'm getting that from this value right here. Now when n is two, what is a sub two? Three. Three, OK. So at two, I'm going to go up to three. So I'm representing this sequence by putting dots in a graph. And when n is three, what is a sub three? Five. Five? OK. So I'll put a dot right there, right across from five. And at four, I think the number would get a seven, et cetera. What do you notice about these dots? Anything special about them? They're forming a line here. Looks like they line up. And it looks like they lie on a line that if I were to graph, if I were to extend the line up, this is the line y equals, what's the slope of that line? Two. So it's 2x plus b. Now let's see, if I'm going up to, then I should be able to find the y-intercept. If I go down two and back one, what number do you think that should be? Negative one. Negative one. y equals 2x minus one. Now this gives me a clue as to how I should calculate a sub n. A sub n is two n minus one. But now when I write it in this form, I'm assuming that I'm restricting my choices of n to integers so that I'm not allowed to pick the decimals and fractions and the other real numbers that are not integers. So by looking at the graph of the line, I can come up with a formula for the sequence. A sub n equals 2n minus one. Let's just check that. When n is one, yes, I get one. When n is two, I get four minus one. I get three, et cetera. So I think that's the nth term that we're looking for. So the answer for part a is that a sub n is equal to 2 times n minus one. That's what we want to define. OK, what about part b? Well, I notice this is an alternating sequence. And also I know that the numbers look like these are powers of two, two to the first, two to the second, two to the third, two to the fourth, et cetera. So if I wanted to find the next term first of all, what would you guess the next term here would be? Negative 64. Negative, right, because the signs are alternating, so it's time for a negative. And if I double this number, so if I multiply by another two, I get a 64. So what would be the nth term in this case? Well, it looks like we need an alternator, so I'm going to put a negative one to the nth power. Now before I complete that, let's just take a look at this nth power. If I substitute n equals 1, then this is negative 1 to the first power. That's going to be negative. But look, we have a positive over here. So what I need to do is adjust this so that I'll get a positive when I plug in a 1. So what I'm going to do is to add a 1 in the exponent. Now what that does is it shifts the pluses and the minuses over 1 space. Now if I substitute in a 1, I get negative 1 to the second power, and I get a positive. OK, and the rest of the sequence we said it looked like was powers of two. So this is 2 to the n. 2 to the first, 2 to the second, 2 to the third. And if I wanted to calculate the, this is the sixth term, if I wanted to calculate a sub 6, that would be negative 1 to the seventh power times 2 to the sixth power. And that's going to be a negative 1 times, now, 2 to the sixth is 64. And so sure enough, I get minus 64. And that agrees with what we guessed was going to be the next number in the sequence. OK, so in part b, we've determined that the nth term, let's call this b sub n, that the nth term is negative 1 to the n plus 1 times 2 to the n. OK, now let's look at part c. Part c, I have 1, 2 fifths, 325ths, 4, 125th, and then suddenly here's 1 over 125th. Let's see if we can find a pattern in this. You know, this first number I could put over a 1, and then look at my numerators, 1, 2, 3, 4. But I don't have a 5 there, but I do have 1, 2, 3, 4. What about these denominators? What do you notice? These are powers of 5, but we started with the 0 power. And then 5 to the first, 5 to the second, 5 to the third, 5 to the fourth. But this doesn't seem to agree with that, does it? What would you expect this denominator to be? Not 125. 625. 625. OK, so what we're expecting right here is a 625 on the bottom, but you know, we were also expecting a different numerator. What were we expecting there? We were expecting a 5. Well, you see what's happened is that 5 will cancel with the extra 5 that I've just put in the denominator, and that reduces to be 1 over 125. This is just the reduced version of that. So I have consecutive integers in the numerator, and I have powers of 5 in these denominators. Now let's see if we can find an expression that would be the nth term. I'll call this one C sub n. OK, well these are fractions. It looks like I have the position number in the numerator, position 1. There's a 1. Position 2, there's a 2. On the bottom, I have powers of 5. Say 5 to the n, that would make powers of 5. But I'll have to adjust this in some way, because I want to start off with 5 to the 0 power here, or 1, in the denominator. So how should I rewrite that exponent? Minus 1, n minus 1. n minus 1, exactly, very good. And I think I'll move that over to get it out of the way. You see, if I want to calculate C sub 1, I plug in a 1 on top and a 1 on bottom, and that'll be 1 over 5 to the 0. 1 over 1 is 1. And if I substitute in a 2 for n, I get 2 over 5 to the 1st, 2 over 5, and I think that will generate all of these terms. So we are guessing what the pattern is, and then we're writing a formula that matches the pattern that we've guessed. But as we saw earlier, the next terms are not necessarily obvious, but I think this is the most obvious thing for us to write down. OK, let's go to the next example and see how sequences come up in other ways besides just lists of numbers. These expressions you might refer to as figurative numbers. It says, for each of the figures that were given below, draw the next figure, and then find the next number in the corresponding sequence, and then find the nth term of that sequence. Well, let's see. Now, you see, here we have an array of dots, and there's a number associated with this array, namely 2. There's a number associated with this one, namely 6, 6 dots. How many dots do I have there? 12, yeah. And here, let's see, it's 5 by 4. Looks like that would be 20. So this is the number sequence associated with those figures. So the first question is to draw the next figure. How many dots, or what should be the dimensions of the array of dots? Yeah, it looks like what we're doing is every time we add an extra column and we add an extra row. And then here I add an extra column, and I add an extra row. So if I had four rows, five columns, the next one's going to be five rows and six columns. So this is the next figure in the sequence. And that's 5 by 6. How many dots are in that? 30. 30. Exactly. OK, so we have done the first thing. We've drawn the next figure. And then we found the next number in the sequence. But finally it says find the nth term of the sequence. Now that's fairly abstract. Can you come up with a formula for the nth term of the sequence? Does anybody see a pattern here that would allow us to write, shall we say, a sub n for these numbers? Stephen, what do you see? n squared minus n. n squared minus n. Tell us how you got that. I think that actually works. Well, I looked at the first term and saw that n squared minus n. OK, now what are you using for n here? That one I actually used too, so I guess it would be n plus 1 squared minus an n plus 1. So what I saw here was that that's kind of a 2 by 2 dot. I think this is the right answer if you make one change in it. I think if you put a plus here, I think it works. Because you see, in the first position, n is 1. And I want to get 2. And this is 1 plus 1 is 2. Now you notice this has one row. It has two columns. The number of rows is the same as the position number. The number of rows here is 1. The number of rows here is 2. The number of rows here is 3. But then you have one more column. So if you're in the nth position, if you're in the nth position, if you could show what I'm writing here, if you're in the nth position, there are n rows and there are n plus 1 columns. You see, in the first position, there's one row. In the second position, there are two rows. So in the nth position, there are n rows. And there are n plus 1 columns. And if you multiply that out, Steven just about had the formula there. It's n squared plus n rather than n squared minus n. So I think this is the formula for the nth term. Let's just check it on, say, the 30 right here. This is in the fifth position. And if I plug in a 5, I get 5 squared plus 5, which is 25 plus 5 is 30. Yeah, so that would give me that number. And I think it gives me all the other numbers as well. OK, now let's look at sequence B. Here's another figurative sequence. In other words, we're not actually given numbers, but we're given some sort of figures. Here we have one triangle, three triangles, five triangles. Now, it's not exactly clear how the number would be arrived at here. What I'm thinking is the number would be the number of sides. I have three sides here. In other words, if we were going to build a set of matchsticks, it would take three matchsticks. How many matchsticks would it take to build this one? Seven. One, two, three, four, five, six, seven. Yeah, it takes seven matchsticks to build these three triangles. How many matchsticks to build this one? One, two, three, four, five, six, seven, eight, nine, 10, 11? I think 11 in that case. OK, now what would the next figure look like? What should I do to that to draw the next figure? Based on the pattern that you see. Just add another four triangles instead of three. Yeah, here I have one triangle, three triangles, five triangles. I think I'm going to have to make seven triangles. I'll add two more triangles. So I think it's going to look like this. I think it's going to look like that. This has two extra triangles. These two triangles have been added on the end here. And how many matchsticks would it take to build that? Fifteen. Fifteen. OK, fifteen. So three, seven, eleven and fifteen. This is the number sequence I have in mind that goes with this figurative sequence. Now can we write the nth term for this sequence? Well, let's see, three, seven, eleven, fifteen. How much are these increasing by? Four. These are increasing by four every time. And so what this makes me think is, I bet a graph would help me represent this. And like we did a moment ago, if I draw my graph, one, two, three, four. At one, I'm going to go up to three, because that was the first number of my sequence. And at two, I'm going to go up to seven. Here's seven. That's seven. And that's three. And at three, I'm going to go up to eleven. And you see, I think these points will again line up like I saw in an example just a moment ago. And the slope of this line is four, because the numbers that we saw going from three to seven to eleven were increasing by fours. So the slope is four, and the y-intercept, anybody guess what the y-intercept is back here? Negative one. Negative one, I haven't drawn it very well, but I think if we go backwards, if you go down four and back one, you'll be at negative one. So this is the line y equals four x minus one. And therefore it suggests that a sub n should be four times n minus one. Now these two formulas represent exactly the same thing. These are actually both functions, except in the second case, I'm assuming the domain, the value of n, is the set of positive integers. As a matter of fact, you could actually say that a sequence is a function from the set of positive integers into the set of real numbers, because you're only allowed to pick integers for n, whereas up here you're allowed to pick any real number for x. So the nth term in part b is a sub n equals four n minus one. Okay, so just to kind of summarize this to the side, I'm gonna say, maybe I should call it b sub n, because this is in part b, b sub n is four n minus one, in that case. Okay, now there are other ways to define sequences besides being given a formula for every n. And one of the alternative ways of defining a sequence is to use what's referred to as a recursive formula. And if we go to the next graphic, we'll see several recursive formulas and how they're used to define a sequence. Okay, now what we've seen so far is that if you're given a formula for the nth term of a sequence, you can use that formula to predict any term in the sequence. There is an alternate way to find terms of a sequence, and this is to use what's called a recursive formula. The idea is this, you start out by knowing the first term and you have a formula that calculates every subsequent term based on the term that came just before it. So to get it started, you have to know the first term or two, and then you have a formula that'll calculate every term after that based on the previous terms. Let's go to the next graphic and we'll see an example of this. It says write the first four terms of each of the following recursive sequences. Now it tells us the first term is one, and then we have this generator, you might call it, that'll tell us how to generate the next term if you know the term that came just before it. For example, let's write down the first four terms of the sequence. Well it tells us right away the first term is one, and after that it says a sub two is two times a sub one plus one. Two times a sub one plus one, that would be two times one plus one is three. So it says if you double the previous term and add one, you get the next term of the sequence. Now if I do that again, if I double the previous term and add one, this would be seven, and if I double that term and add one, I get 15. What would the next number of this sequence be if I wanted to get a fifth term? 31. It'd be 31. Now you notice in this case, we don't have a formula that tells us what a sub n is. So if I were to ask you, what is the 50th term of the sequence, there's no way I could get it directly without writing all these terms out. But I do have a formula for a sub n, but it's one that's given recursively as they say, that is it's defined in terms of the previous term. Okay, let's go to part b. Find the first four terms of this recursive sequence. b sub one is negative one, and b sub two is one over b sub n minus one plus two. Okay, so I'm gonna write down the first four terms, and we said the first term here is negative one. Now what would be b sub two? Well, it says take one over the previous term plus two. Okay, what is the previous term plus two? The previous term here plus two is how much? One, and then if I take one over that, I get a one. Okay, now if b sub two is one, what's b sub three? One over three. One over three, exactly. So if I put in a one right here, I get one over three. And if I put in a one third right there, we'll have to come over here and work this out. One third plus two, let's see, one third plus two is seven thirds. Now this says one over that. So what's one over seven thirds? Three sevenths. So this would be three sevenths here. And we could continue to write out more terms, but I can't name a specific term down the line without working out every term in between. Okay, and finally, we have a recursive sequence where the first two terms are given. And I'm gonna work this one out on the green screen because it's gonna take a little bit more space. It says we're given, if we go back to the green screen. Okay, so we're given that c one is equal to one. C two is equal to three. And c sub n is equal to one plus c sub n minus two. Now this is for n greater than two, because we already know what the first two terms are. So if we wanted to find the first four terms, well let's see, the first term is one, the second term is three. And now for every term after that, c sub three, it's one plus the sum of the two previous terms. Well, that would be one plus one plus three is five. And now to find the term after that, c sub four, it's one plus c three plus c two. C three plus c two is eight plus one, this would be a nine. Can anyone tell me what the next term would be after that? 15. It'd be 15, yeah. Stephen, tell us how you got that. Nine plus five plus one. Exactly, what you do is you add up the two previous terms plus one, so that makes 15, very good. And we could continue out for a while, writing out more terms, but I think you have the idea. There's a very famous example of a recursive sequence, and this is known as the Fibonacci sequence. Let's go to the next graphic and take a look at this. In the Fibonacci sequence, the first two numbers start off with ones. And every number after that is the sum of the two previous numbers before it. So the number that goes here is one plus one, so I put two. The number that goes here is the sum of the two previous numbers, one and two is three. And the number that goes here is the sum of the two previous numbers, two and three. Now, this is referred to as the Fibonacci sequence. It's named for a mathematician who lived around 1200 A.D. in Italy, named Fibonacci. Now, let's see if we can define the sequence recursively. I'll need to know what is C1. I'll need to know what is C2. And I'll need to know a formula for C sub n. This is where n is bigger than two. So what is C1 equal to? One. Is one, mm-hmm. What is C2 equal to? One. Is one again. And what is C sub n for n bigger than two? C sub n minus one. Uh-huh. Plus C sub n minus two. Plus C sub n minus two, exactly. C sub n is the sum of the two previous terms. This is how I would represent it. Now, you know, there are physical models of the Fibonacci sequence. And if you come to the green screen, let me just show you a physical model of this sequence. Suppose you have a plant that grows from a seed below the ground. And let's say in the first week, it grows up to this height. And then in the next week, it grows up to this height. But in the third week, it grows up the same amount of distance, but it's now mature enough that it has an offspring or a branch that comes off of it. So once it's entered to the third week, it's now mature enough that it puts off a branch. Okay, now after that, it continues to grow the same distance again, and the branch continues to grow. And then in this week, the main branch is mature enough that it still puts off another branch. It put off a branch here, and it puts off a branch there. Now in the next week, this one grows again, and it's mature enough still that it puts off another branch. This one continues to grow, and now it's into its third week, so it's mature enough to put off a branch over here. This one grows further, but it's not yet mature enough to put off another branch. Okay, now I want you to tell me what happens in the next week. Will this one grow? Yes, will it put off a branch? Well, actually, see, it's putting off a branch every week because this is the main trunk, so it's mature enough to put off a branch. What about this one, will it grow? Yeah, will it put off a branch? No, because it's only two weeks old. You see, you have to be into the third week before we put off a branch. How about this one right here? It grows, should it put off a branch? Yes, it will, right. And this one continues to grow, but it's not mature enough to put off a branch. Back over here, this one grows, will it put off a branch? Yes, it will. Okay, now I think you have a sense of the idea of what's happening here. Look at how many branches and trunks there are every week. Originally, there's one, and then there's one, and then there's two. How many branches are there here? Three. Three. And then there are five. And then there are, I think if you count those, I think you get eight. One, one, two, three, five, eight. This is the Fibonacci sequence, and it can be modeled by the growth of plants. Now, you know, just yesterday, I was mowing the front yard, and there's a tree that I planted three years ago, and in the first year, it grew this much, and in the second year, it grew this much, and I'm not kidding you, this year, in the third year, it put out a second branch. And actually, I'm starting into the fourth year now, and I'm wondering if one of these branches will continue growing and put out another branch, and if this one is gonna grow a little bit more, I'll just have to wait and see what happens later this summer, what's gonna happen with that tree. But my tree is growing exactly according to the Fibonacci sequence, except the things aren't branching every week, but they're seemingly branching in years rather than weeks. Pardon me? Okay, I'll keep you posted. We'll be back in a year and I'll let you know. Okay, let's go on to the next graphic, and we're gonna shift gears now from sequences to series. Now, the difference between a series and a sequence is that in a series, the numbers are added together rather than just listed. So if I say A1 plus A2 plus A3, et cetera, this is a series. There's an abbreviation for this notation. It's called sigma notation, and if I put a capital sigma right here, sigma is a Greek letter. What is it comparable to in our alphabet? We use the Roman alphabet. What's this comparable to a sigma? S. An S, yeah, and S for some, but rather than putting an S here, the notation is usually to put a sigma, and in general, I'll say this is A sub K. You see that I have subscripts. A sub K, where K goes from one to five. So above and below the summation, you tell where do the subscripts begin and where do they end. You pick a letter to represent the subscript. This is referred to as an index. I use the letter K, and K goes from one to five. So this is an abbreviation for that summation. Now, let's look at an example of that. Suppose I said sigma of I squared as I goes from one to four. I'm using an I here rather than a K, but the choice of the letter is really immaterial. Well, this is the sum of the first four terms. I'll call it S sub four. And the summation of the first four terms, I squared as I goes from one to four, that would be one squared plus two squared plus three squared plus four squared. Now you notice I'm putting pluses in here because this is a summation. And if I add those up, if I add up those squares, I'll get 30. Okay, now in the last example that I have down here, this is referred to as an infinite series, and you notice the summation goes from K equals one to infinity. This is the symbol for infinity up here. If I make it bigger, it's a symbol like that. Okay, so this goes from one to infinity, and what that means is if my terms are called A sub K, it's A one plus A two plus A three, and it goes on forever. So that's called an infinite series. Let's go to the next example, and we're asked to evaluate some series. Okay, in the first case, I have the summation of three I plus one where I goes from one to six. Now what does that mean? Well, if I substitute in a one, how much will that be? That'll be a four plus. Now if I substitute in a two, how much will that be? Seven, yeah. If I substitute in a three, Susan, if I plug in a three, how much will that be? No, see if I plug in a three right there, three times three plus one is 10. That'll be a 10. And if I continue to plug in the numbers from one to six, I'll get 13 plus 16 plus 19. Now if I add those up, that's the value of this summation. You know, when I add them up, if I add them up in the right way, I can make this fairly easy. For example, seven and 13 is 20. 16 and 19 is 35. So what do I have here? 35, 55, 65, 69. That'll be 69. So what I've done is I've tried to add them so that I could get groups of tens and fives to make it easier to add. Let's take this next one. This one says, what's the summation of log base two of K where K goes from one to five? Well, I plug in K equals one, K equals two, K equals three, and I add them up. That's gonna be log base two of one plus log base two of two plus log base, whoops, base two of three, plus log base two of four plus log base two of five. Now you might say, Dennis, how are we gonna add those up? Or you might say, Dennis, I know how to add those up using properties of logarithms. We haven't talked about this for a few episodes. First of all, what is the log base two of one? Zero. That's zero. You remember, a logarithm is, there's a famous statement about that. A logarithm is what? An exponent. A logarithm is an exponent. It's the exponent you put on this base to give you this answer. Well, two to the zero is one, so this is zero. Now, what about these other logarithms? This one I think we could actually evaluate, but I'm looking at this from a different point of view. When you add logarithms together, you get the log of the product. So this will be two times three times four times five. That's the log of the product. And that will give me the log base two of, let's see, here's a 10 and a 12. 10 times 12 is 120. The log base two of 120. That's the summation of this series back here. I just stuck this one in so we'd have to review properties of logarithms along the way. I thought it was a good opportunity. Okay, now we have one more series given down at the bottom. And let me write this one on the screen. That last one that we didn't really have room to write on went like this. It said we were supposed to find S1, S2, S3, and S4 for this series. And the series was I plus one over I. I goes from one to N. Now let me explain why I'm putting it in here. I didn't want to say specifically where this series ended. It starts at one and it goes to some integer N. So it doesn't really matter what that N is, but this is the formula for each term. What is S1? Well, not S1 is the sum of the first term. What is the first term of that series? Two. Yeah, if I plug in a one, I get one plus one over one is two. So when I say the sum of the first term, that's actually kind of an abuse of the word sum, I guess I'm not really adding anything, it's just the first term. S2 is the sum of the first two terms. That would be one plus one over one plus two plus one over two. That would be, we already calculated that, that would be two plus three halves. Which is seven halves. Now what is S3? Well, S3 would be the sum of the first three terms. Well, you know we already know the sum of the first two terms. So why don't we just add on one more term to that? What would be the next term I'd add on to this? What would be the third term? Four thirds. Four thirds. So what I'm gonna do is add four thirds to the sum of the first two terms. This is the first two terms already added together. And I think we should get a common denominator of six. So this would be 21 sixth plus eight sixth is 29 sixth. Okay, finally what is S sub four? Well, technically S sub four is the sum of the first four terms. But I already know the sum of the first three terms is 29 sixth. And what should I add to that? What should I add on? Five over four. Yeah, so what we're doing is plugging into four right here, five over four to get the fourth term. Five over four. The common denominator there, the least common denominator is 12. That would be 58 twelfths plus 15 twelfths is 73 twelfths. Okay, so here's what we got. S one is two. That's just the first term. S two is seven halves. That's the sum of the first two terms. S three is 29 sixths. That's almost five. That's the sum of the first three terms. And S four is 73 twelfths. That's a little bit more than six. Little over six. Okay, now we've had a little bit of experience looking at this sigma notation. I think we should look at some more examples where we have to introduce it. Let's go to the next graphic and I'll show you what I mean. And this example says write the following using summation notation. Okay, so we've got three plus six plus nine plus 12 plus 15. Now there are five terms there. So I'm thinking we should have this go from i equals one to five. And if you don't want to use i, you could use j or k. Those are the most common letters used. But basically each one of these terms is triple the position number. Like in the first position, it's tripled, I get a three. In the second position, it's been tripled, I get a six. So I think this should be three i. On the other hand, if you put a k here, if you say k goes from one to n, then I'd put a k over here. You see, the choice of the letter doesn't matter. So whatever letter you pick here, you just put over there. Sometimes it's referred to in mathematics as a dummy variable because the choice is immaterial. It's just what it represents that's significant. Okay, let's look at the summation in the next line. You notice this is not a sequence because I have pluses in between, so I'm adding these up. If I want to abbreviate that, it would be a summation. What if we use a j here? Let's say j goes from one, one to what? What would you guess would be the upper number? Four. One to four, because I have four terms. So I should have four choices for j. And it looks like these are all fractions with numerator one. So I'll put a one on top. Now, one times two, two times three, three times four. What do you get out of that? J squared plus j. J squared plus j. What if you put that in factored form? How would you write it? You're actually right, but let's put it in factored form for everyone else. J times j plus one. J times j plus one. Yeah, that's, I think, easier for everybody to see. Because what we're doing is we're taking the position number times the next number, the position number two times the next number, and the position number times the next number. So j times j plus one. So this is an abbreviation for that. Now, you might say, well, Dennis, you know it's not actually that difficult to write out those four terms. Do we need to know the summation notation? Oh, most definitely, because what if there weren't four terms, but maybe 400 terms written out? All I would do to change this to 400 terms is I would have the scope to 400, and that would represent a much, much longer series. So it's important to know this abbreviation. You might say, well, now, in what courses do you use sequences in series? Well, if you're gonna go on to take calculus and differential equations, say if you're a math or engineering or computer science major, if you're gonna take courses in statistics and probability you study or you make use of sequences and series, so this information is beneficial to many disciplines. Okay, we have one more summation given in part C, and we have one minus one over x plus one over x squared. Now, you notice we have this alternating sign, so I would call this an alternating series, an alternating series, and the first thing I need to do in my series is put in an alternator. Now, let's say we're gonna have K be the index. K goes from, now, let's say we go from one to six, because I have six terms. I don't think I'm gonna stick with that one, but I'm gonna put one there for the time being. Now, if I put in an alternator, that's gonna be negative one to some power, we'll come back and figure out that power in a moment, times one over x to some power here. Looks like this is one over x, one over x quantity squared, one over x quantity cubed. Now, you notice that the very first term doesn't have a one over x. So I think we should actually begin our series starting off at zero so that my exponent, if I call it K, that I can let that be zero, and I'll get a one right there. Also, if K is zero, I want the first term to be positive. So if I put a K right here, K to the negative one to the zero power will be a positive, and that'll be one times one is one. So I believe this will take care of all of those terms, and this is an abbreviation for it. You know, another way to write that, if I combine it, I'll just put it right up above here, is to say the summation of negative one over x to the K power, where K goes from one to six. That is, what I've done is to multiply those together and put one exponent outside. You could write these either way. Stephen. Shouldn't it be zero to five on both of those summations? Oh, yeah, you're right, zero to five. Yes, exactly, thank you. When I change that lower bound, I should have changed my upper bound. That would be five, and this one I should say from zero to five also. Thanks for pointing that out. If you're gonna adjust the lower index, you should adjust the upper index with it. Okay, let's go to the last graphic, and this one says some sums aren't so hard to sum sometimes. Okay, if you catch my drift. And the idea here is to show a shortcut for computing the sum of a series. Now in this first case, if I write out the terms, let's see, if I plug in a three, that's gonna be 10 to the third minus 10 squared. And the next term, if I plug in a four, will be 10 to the fourth minus 10 cubed. And then if I plug in a five, I get 10 to the fifth minus 10 to the fourth. And the last term, I'm running out of room there, the last term is 10 to the sixth minus 10 to the fifth. Now if you look at what's happening overall, there's a negative 10 cubed and a positive 10 cubed. Those cancel out. And there's a negative 10 to the fourth and a positive 10 to the fourth. And there's a negative 10 to the fifth plus a positive 10 to the fourth. This is sometimes referred to as a telescoping series because alternate terms seem to cancel out and what I'm left with is only two terms. I have, I'll go down to this space. I have, looks like 10 to the sixth minus 10 squared. Now 10 to the sixth is a million and 10 squared is 100. A million minus 100 is 999,990. And what I've been able to do is to find the sum of the series without multiplying all those out but by canceling the terms using this idea of a telescoping series. Now not all series have this property but some do and it's something to keep in mind because it comes up in the homework. Okay, let's look at the next summation. I wanna add one plus three plus five, et cetera. I wanna add up all these numbers. So the way I'm gonna add them is I'm gonna group nice terms together. How about 99 and one? That makes a thousand. And then 97 and, oops, and three. Let me come all the way over to three. That's a thousand. And if I pair these numbers up, the smallest and the largest, I get thousands. Now the question is how many thousands would there be? Well you know, what I've listed are the odd numbers from one up to a thousand. So there are 500 odd numbers listed here. So I have 500, I have 500 thousands. So when I add these numbers up, what I get is 500,000. And the way I did it is I matched the biggest and the smallest, then the next biggest and the next smallest, and I saw that every time it was a thousand and I added up all those thousands. I don't think we have time to do the same thing in the last example, but if you look on the website, you'll see the last example worked out as well. Okay, so today we've had a chance to introduce sequences in series. In the next episode, we'll look at two special sequences in series, that is arithmetic sequences and geometric sequences in series. I'll see you next time for episode 31.