Worked examples: finite geometric series | High School Math | Khan Academy

So we're asked to find the sum of the first 50 terms of this series, and you might immediately recognize that it is a geometric series. When we go from one term to the next, what are we doing? Well, we're multiplying by ( \frac{10}{11} ). To go from 1 to ( \frac{10}{11} ), you multiply by ( \frac{10}{11} ). Then you multiply by ( \frac{10}{11} ) again, and we keep doing this.

We want to find the first 50 terms of it, so we can apply the formula we derived for the sum of a finite geometric series. That tells us that the sum of, let's say in this case, the first 50 terms, actually let me do it down here.

The sum, the sum of the first 50 terms is going to be equal to the first term, which is 1. So, it's going to be ( 1 \times (1 - ( \frac{10}{11})^{50}) ) over ( 1 - \frac{10}{11} ).

I'm not going to solve it completely, but we can simplify this a little bit. This is going to be ( 1 - ( \frac{10}{11})^{50} ) over ( \frac{1}{11} ). So, this is the same thing as multiplying the numerator by 11.

So this is going to be equal to ( 11 \times (1 - ( \frac{10}{11})^{50}) ). You could try to simplify this even more, but this gets us pretty far. At this point, it is just arithmetic.

Let's do another one of these; this is kind of fun. So this is more clearly a geometric series. Let's just first think about how many terms we're going to take the sum of. You might be tempted to say, "Okay, I'm taking to the 79th power; there must be 79 terms here." But be very careful, because the first term is when we're taking things to the zeroth power. We're taking ( 0.99 ) to the zeroth power.

The second term is where we're taking it to the first power. The third term is where we're taking it to the second power. The fourth term is where we're taking it to the third power, and so on and so forth. So this right over here is the 80th term, the 80th term. So we want to find ( S_{80} ), and so this is going to be equal to our first term, which is going to be ( 1 \times (1 - (-0.99)^{80}) ) all over ( 1 - (-0.99) ).

At first, you might say, "Well, maybe the common ratio here is ( 0.99 )," but notice we have a change in sign here. The key thing is to say, "Well, to go from one term to the next, what are we multiplying by?" Well, to go from the first term to the second term, we multiply by ( -0.99 ). Then, to go to the next term, we're again multiplying by ( -0.99 ).

So the common ratio is not positive ( 0.99 ), but negative ( 0.99 ). So let me write that: ( -0.99 ), and of course that is going to be to the 80th power all over ( 1 + 0.99 ).

We could simplify this a little bit. This is all going to be equal to, well, that one we don't have to worry too much about. So this is going to be ( 1 - (-0.99)^{80} ). Actually, put parentheses there to make sure we are taking ( (-0.99)^{80} ).

Well, we're taking it to an even power, so it's going to be positive. So that's going to be the same thing as ( 0.99^{80} ), and all of that over. Well, subtracting a negative that's just going to be adding the positive, so all of that over ( 1.99 ).

We could attempt to simplify it more, but with a calculator, we could actually find this exact value or a close value. Actually, most calculators don't give you the exact value when you take something to the 80th power, but this is what that sum is going to be.

Let's do one more of these. All right, so here we have a series defined recursively, and so it's useful to just think about what it would actually look like. The first term is 10, and then the next term, so the second term ( a_2 ) is equal to ( a_1 \times \frac{9}{10} ).

Right, so the next term is going to be the previous term times ( \frac{9}{10} ). So it's going to be ( 10 \times \frac{9}{10} ), and then the next term is going to be that times, is going to be the second term. The third term is the second term ( \times \frac{9}{10} ). So ( 10 \times \left(\frac{9}{10}\right)^2 ).

The way it's written right now, I haven't written it as a finite geometric series, so let's say we want to take the sum. Let's say we want the sum of the first ( 30 ) terms. The sum of the first ( 30 ) terms is going to be equal to the first term. We've done this before, the first term times ( (1 - r^{30}) ) over ( (1 - r) ).

Let's see, we could ( 1 - \frac{9}{10} ). This is ( \frac{1}{10} ) right over here. You divide by ( 1 ); that's the same thing as multiplying by 10. So this is going to be ( 10 \times \left(1 - \left(\frac{9}{10}\right)^{30}\right) ).

Actually, these parentheses, you always want to, so I put parentheses there just to make sure we see that we're taking both the 9 and the 10, or the ( \frac{9}{10} ) to the 30th power—not just the 9.

So there you go. Did I? Yep, there you go; we're done.