Common Core Algebra II.Unit 5.Lesson 5.Geometric Series

Hello, my name is Kirk Weiler, and this is common core algebra two. By E math instruction. Today, we're going to be working on unit 5, lesson number 5, on geometric series. Yesterday, we introduced you to an arithmetic series. Which was based on arithmetic sequence, and today we're going to do the same thing. We're going to work with geometric series, which will of course be based on geometric sequences. So let's jump right into it and start to see how we can add up the terms of a geometric sequence to create a geometric series. All right. Plain and simple, right? The geometric series, the sum of the first N terms of an of an arithmetic series no, no, no. That's supposed to be the sum of the first N terms of. G metric series sequence. We're going to have to talk to the editors on that. So let's take a look, right? Given a geometric series defined by their cursive formula a one equals three and a N equals N minus one times two, which of the following is the value of this sum. We'll play around with this a little bit. You have all the tools you need to answer this question. So use this, pause the video and see where you're at just as we begin this lesson, okay? All right. Well, let's talk about it. Again, I hope that you look at this at this point, and I know it is kind of Greek character in it, subscripts, indices, and all sorts of things. But all this is, is the sum of the first 5 terms. That's it. So now we have to generate those terms. And how do we generate them? Well, we know a one. That's three. And then we know that each term is twice the previous term. That's what that recursive rule is telling us. Each term is twice the previous term. So that's going to be simple a two is 6. A three is 12. A four is 24. A 5 is 48. And when I add them all up, that's what that sigma means. I'm going to get what? Gotta look at my cheat sheet, 93. That's it. So again, this problem was more about being able to interpret a recursive definition for a geometric sequence, and then also interpret the sigma or summation notation to simply asking us to do the following. All right. Pause the video now and write down anything you need to, and then we'll move on to the first real problem here. All right, let's do it. Now, if you recall, in the last lesson, when we worked with arithmetic series, we were able to exploit a pattern with the arithmetic sums where they would pair off in a constant sum. I wished the same thing happened with geometric, but it's simply doesn't. All right? Still, we can definitely use structure and strangely enough things that we know about factoring and exponents to be able to come up with a really nifty formula for summing or evaluating and a geometric series. All right, now, before we do that though, I have to review something with you, which is going to be very, very important. How we calculate the n-th term of a geometric sequence, not series, sequence, right? Remember, we always do that by taking the first term and multiplying it by the common ratio, whatever we're multiplying it by one less time. All right? So in other words, what we have when we have an arithmetic, I'm sorry, a geometric series sequence, is we have our first term, then we have our first term times R, then we have our first term times R squared, then we have our first term times R cubed. Et cetera, all the way up to our last term, but our last term isn't our first term times R to the N, it's times R to the N minus one. There's our geometric sequence. When we add them all together, here is our geometric series. And what we're going to do in this exercise, because I really want to be very upfront about this, is we're going to find an easy formula to evaluate the sum. We're going to be looking for a formula that evaluates that sum. Okay? Now, the first thing that we're going to do is we're going to strangely enough take that sum and multiply the entire thing by R all right? So we're going to get R times a one plus a one R plus a one R squared loves a one R to the N minus two plus a one R to the N minus one. It's very important that we this is our last term. So what do you get when you multiply through when you distribute the R? Think about that for a minute. Write down what you think you get. It's not that hard, right? You'd get a one times R and that would be our first term. Then a one times R times R will be a one R squared. And of course, this one would become a one R cubed. In fact, all these powers are just going to go up by one. Which means this power is going to now be a one R to the N minus one. And this one's going to be a one R to the N that's very important. Okay. Now letter B says find and simplest form, the value of SN minus R times S so I'm going to take this thing and I'm going to subtract off this one. Now what I'd ask you to do, I'm going to circle this in red, and I'm going to circle this in red. And I'm going to subtract this one from this one. Now, something amazing happens. I want you to think about what would happen if you subtracted this kind of polynomial right from this one. What would happen? Pause the video now and really take a look at that. Well, I do want to point out something. They have the same number of terms. They have N terms. Now, you start subtracting, you think Matt, no big deal. A one, I don't know, that can't do much with that, right? But then if I take this and I subtract this, they cancel. If I take this and subtract this, they cancel. And of course, there would be one here that we cancel this, et cetera, et cetera. There would be one right before this one that would cancel that one. And then this one would cancel that one. But then there would be nothing to subtract this. So it would just go on. That's that. It's amazing. Every single term in those two sums are identical except for this term, which doesn't occur down here. And this term, which doesn't occur up here. So when we subtract all those middle terms cancel out and we're just left with these two. Which is amazing. Now here's the real key though. Letter C says write both sides. I'm going to go back to blue. Write both sides of this equation in its factored form. So I'm going to write this down. It's a one. I'd like you to factor this and factor this. Pause the video really quickly and do that. All right, well, given that we just did a bunch of factoring in the last unit, it shouldn't be that big of a deal in both cases. Their GCF factors. So here I can factor out an S sub N, which will leave me with a one on this term. And a minus R on that term, here I can just factor out an a sub one, which will leave me with a one. And then an R to the N all right, that's actually all the problem asked me to do. Just factor. But now we can find a formula for S sub N and we can find that formula by simply dividing by one minus R on both sides. And we now have a formula for our geometric series sum. There it is. Now let's before we move on, talk a little bit about this formula. Actually, maybe we'll talk about it on the next sheet. Pause the video now and think about what we just did. We used a lot of structure. In order to find this formula, so pause a little bit and think about it and then we'll talk about the formula more on the next sheet, okay? All right, let me clear this out. And we'll take a look. Or we won't. Let me write down the formula now. All right. And then we'll use it. And then we'll talk about it a little bit more. All right. So the formula is actually pretty pretty easy. But it requires different things that are arithmetic series formula. It requires the first term. It requires the common ratio. Ratio. Which is that R term. And it requires the number of terms. But as long as you know all of those, then you can pretty easily do this. So it says which of the following represents the sum of a geometric series with a terms. So that's N, whose first term is three. That's a one. And whose common ratio is four. That's your R, right? So I can say, well, my 8th sum is going to be my first term three times one minus my ratio for raised to the N it's a, all divided by one minus four. All right, now again, that formula or that evaluation is not that difficult to do. But pause for a minute, type all that into your calculator, and then I'll give you the answer just to make sure that we can all do this on our calculator. All right. Well, at the end of the day, at the end of the day, I typically evaluate the numerator first and then I divide by whatever the denominator is going to be, but it is huge. Geometric sequences with common ratios greater than one. Grow very quickly. So when we add them all together in a geometric series, they tend to get very, very large. Now again, if that common ratio is between zero and one, if it's actually making the sequence go down, think about exponential functions. That's what geometric sequences really are. Exponential functions. If they're going down well, then the sum doesn't end up necessarily being all that large. But when that geometric series or sequence is going up, because the ratio is greater than one, here the ratios four, then the sums can get very large, very fast. All right, pause the video now, and write down anything you need to. Okay, let me clear this out. All right, so here's the sum of a geometric series. And again, let's make sure you really get it. Because it's different, right? The sum of an arithmetic series was this. And for an arithmetic, what we needed was the first term, the last term, and the number of terms. For a geometric, we need the first term and the number of terms. But we need the common ratio instead of the last term. Okay? So it's good to have that formula just sitting in front of us, sitting there. Sometimes it's written like this where that first term has been distributed in the numerator. It doesn't matter. It really doesn't either way. I think this is an easier calculation, but that's just me. So let me clear all that out. Let's do exercise four. Find the value of the geometric series shown below. Show the calculations that lead to your final answer. Well, okay, fine. What do we need? We need the first term. And we have it. That's 6. We knew that the common ratio. What is that? Ah, it's two, right? We're multiplying by two, we're multiplying by two, so common ratio is two. And then we need the number of terms. So that's not 768, right? That's our last term. We need the number of terms. All right. Well, we are going to pull a maneuver that is very, very similar to what we did with our arithmetic series. What we're going to do is we're going to use this fact. The last term is always equal to the first term times the common ratio to the N minus one. But here, we're trying to figure out the N, right? We know that the last term is 768. We know that the first term is 6, and we know that the common ratio is two. What we don't know is N so we're actually, this is really kind of cool. We're going to solve an exponential equation here. Now, be careful. The first thing I'm going to do is divide both sides by 6, all right? I'm going to put that two to the N minus one on this side. But if we do that division, sorry, let me just kind of get here. Then what we'll find out is that we get 128. We could use logarithms to solve for N or we could use the method of common basis, which is what I think I'm going to do. 128 is two to the 7th. All right, two to the 7th is 128. So N minus one is 7. And we have a terms. We got everything now, right? So finally, the 8th sum is going to be my first term. Times one minus my ratio, which is two, raised to the number of terms I have 8 all divided by one minus two. Let me just do that. All right, I have to be very careful when I enter that in my calculator because there's a lot of parentheses. There's big numbers. There's negatives. But I find 1000 530. So it's not an overly difficult formula to use. But it's not nearly in my personal opinion as intuitive as the arithmetic series, where it's to me very obvious how those sums pair off and why we have N divided by two of them, et cetera. This is a formula that definitely came from the structure of a geometric sequence and then by extension of geometric series. But it is not nearly as intuitive. I don't look at this formula, either one of those two at the top of the screen and go, oh, of course. Yeah, no, of course, it's a one times one minus R to the end divided by one minus R that's just not obvious to me. And I've been doing this for a long time. So pause the video now and then clear out anything you need to. All right, let's clear it on out. Here we go. So let's move on to a very practical application. One of the really neat applications for geometric series. Is when you deposit money into an account that earns some kind of an interest, right? We've done that many, many times. You deposit money into an account that earns interest, but you keep making deposits. So up to this point, right? You know how this works. I deposit $500 and then I never touch the account again. What we're going to do is look at a situation where Maria places $500 at the beginning of every year into an account that earns 5% interest compounded annually. But she's going to keep putting the $500 in there at the beginning of each year. So she'll put $500 in there January 1st. And then the next year, she'll put $500 in there, and then the next year she'll put $500 in there, et cetera. So just take a look at letter a, it says determine a formula for the amount a of T that a given $500 has grown to T years after it was placed into the account. This should be very, very quick and easy. Why don't you go ahead and do this? Sounds like there's a B buzzing outside. It's actually somebody who's cutting down trees, but what are you going to do? The videos must go on. I would love to claim that I'm in Los Angeles, California, or something inside of a movie studio B or something, but that's just not true. So here is the amount of course that a given $500 deposit would be worth two years after I've put it into the account or after Maria has. So let's take a look at letter B it says at the end of ten years, which will be worth more. The $500 invested in the first year or the fourth year. Explained by showing how much each is worth at the beginning of the 11th year at the beginning of the 11th year. So intuitively, you probably realize that the $500 that we invested in the first year is going to be worth more than the $500 invested at the fourth year. But how do we see that? Well, the $500 that we put in right at the first year sits there for ten years. So it's worth 500 times 1.05 to the tenth. And if you work that out, it's worth $814 and 45 cents roughly. So this is the money I put in January 1st on the first year. On the other hand, the money I put in on the fourth year, right? It only got to sit in there for 6 years. And that ends up being $670. And 5 cents. So clearly this money has been sitting there longer. But what we really want to know is how much Maria has in there. And to make the sum a little bit more a little bit less confusing, sorry. What we're talking about is putting money in there, including money that she's going to throw in right at the beginning of the 11th year. So let's add up all the money. Well, you got the money she put in there the first year, right? And it's going to be worth 500 times 1.05 to the tenth. Then you have the money she put in there the second year. And that's going to be worth only 500 times 1.05 to the 9th, right? Then the money she put in there the third year, 500 times 1.05 to the 8th. Plus, et cetera, all the way down to 500 that she put in there, right? Let's put in one of these to the first, and then the money she deposited right at the end. Oh, well, we don't need to put any of that in there. There we go. Right now, normally, we write the geometric series the other way around. But this is kind of the way I was thinking about it, right? This is the money we put in there right at the beginning of the first year. And then the second year, et cetera. Now, of course, there's not that many terms. We could actually just add these up by hand. But let's take a look at how the formula works, right? What was that? It was take the first term. That's actually going to be this guy. Multiply it by one minus R to the N all divided by one minus R now we have actually have everything we need, but we have to be a little bit careful. The first term is 500. All right? The common ratio is the 1.05. But how many terms do we have? You might think we only have ten terms, but we actually have 11, right? Because we have this one. This is our first term. That's our second term. Et cetera, right? We actually have 11 terms. Think about that a little bit. Okay, I'm going to write all this calculation down. But I really want you to think about why we have 11 terms. Give that for a second. But then once we have this, it's just a matter of pushing the buttons on the calculator. And we end up with 7000 $103. And 39 cents. Which is rather cool, and it's a lot more realistic than the typical investment scenario that you see even with the compound interest that we did a unit ago. This is way more realistic. The idea that I would put money in and it would sit there and gain interest, I would put some more money in. It would sit there and gain interest, but less than the first amount of money I put in, right? That is a much more typical investment scheme than dropping some money into an account walking away in 20 years later, pulling it back out. Yeah, some people do that. But most of the time, there's this sort of defined, I'm putting money in at a regular interval kind of thing. I have money taken out of every paycheck and it's put into a retirement account. There's some money that's taken out of every one of my paychecks and put into my children's college savings accounts. And it's always the same amount of money. Now, whether or not I'm always getting 5% interest, that would be nice. Nobody gets 5% interest these days. No matter what interest I get, whether or not that's consistent, you know, whether I'm always getting 5% or always getting 3%, that's more of a question. But I digress. Pause the video now and write down anything you need to. All right, let's clear out the text and do one more applied problem. Here we go. A person places one penny in a piggy bank on the first day of the month, two pennies on the second day, four pennies on the third day, and so on. Will this person be a millionaire at the end of a 31 day month? Show the calculations that lead to your answer. Wow. Okay. So one penny and a piggy bank on the first day. Two pennies on the second day, I'm doing this in terms of dollars. Four pennies on the third day. 8 pennies on the fourth day. I don't know what, on the last day. And there we have it. Okay, so the question is, will they be a millionaire? Well, pause the video now and see if you can figure this out. At first glance, it doesn't seem like they're going to. I mean, one penny, the first day, two pennies, the second, four pennies, the third, 8 pennies the fourth, 16, the 5th. I mean, 16 cents on the 5th day, how am I going to become a millionaire? Well, let's take a look at that sum. This is definitely a geometric series, right? A one times one minus R to the N, all divided by one minus R well, we have everything. We have a one. We have R it's two. And we have the number of days. It's 31. So let's do it. .01 times one minus two to the 31st. All divided by one minus two, and if you work that out, you're more than a millionaire. 21 million, 474,000. $836. And 47 cents. So unequivocally, yes. That person can is definitely going to be a millionaire. I don't know where they're going to get all that money to deposit in the piggy bank. I also don't know where they're going to get a piggy bank that's big enough to hold $21 million. Actually, almost 21 and a half $1 million. But hey, that wasn't the question. The question was just whether or not they'd be a millionaire. All right. Pause video now, write down anything you need to. All right, let's clear out the text. And wrap up the lesson. So today, we saw geometric series. In the last lesson, we saw arithmetic series. The geometric series formula is much more complicated than the arithmetic one. But still, you know, if you're comfortable with everything that it requires, IE the first term, the common ratio, and the number of terms, then you can simply use it just like you use that arithmetic series formula to add up all the terms of a geometric sequence. In the next lesson, we're going to look to see how this applies to mortgage payments, which is quite challenging. So make sure that you work hard on the homework on here, and that you really learn that formula because in the next lesson, we're going to take it up a notch. All right. For now, I want to thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking and keep solving problems.