 Suppose we have a geometric sequence a-n is equal to a times r to the n minus 1. Suppose we have this geometric sequence. And let's say we want to add together the first n terms of the geometric sequence. That is, we want to look for the partial sum s sub n here. Well, by the formula, we're going to see that s sub n, right, is going to equal the first term, which is a, plus the second term, which is a-r, plus the third term, which is a-r squared, all the way up to the nth term, which will be a-r to the n minus 1, like we saw before. Now, if we take this sequence, this partial sum right here, and we times everything by r, right, r times s-n, well, we can distribute that r into all the terms of the sequence, and we're going to end up with an a-r plus an a-r squared plus an a-r cubed all the way up to an a-r to the n. And so we get this from just times in s-n by a-r. What we're going to do next is we're going to subtract these things. If we take s-n minus r-s-n, this is going to look like a plus a-r plus a-r squared all the way down to a-r to the n minus 1. That's the s-n part. And then if we subtract from it the r-s-n part, we're going to get an a-r plus an a-r squared all the way down to, we're going to get an a-r in minus 1 and a-r to the n, like so. And so on the left-hand side, you have s-n minus r-s-n. Notice there's a common factor of s-n there. We can factor it out, and we're going to end up with 1 minus r times s-n. That's the left-hand side. On the right-hand side, you're going to notice there's a lot of cancellation going on here. You're going to have an a-r minus an a-r. They will cancel out. You'll get an a-r squared that cancels with that. Then the a-r cubes will cancel, and this will continue all the way down until you get to a-r in minus 1, cancel a-r in minus 1. This will leave us with just two terms really, a term at the very beginning, which is a, and then you're going to have this negative a-r to the n at the end there. In which case if we put those together, we're going to get a minus a-r to the n, and in which case then to solve for s-n, what we can do is we just divide both sides by 1 minus r, divide by 1 minus r, and in which case we then get that s-n will equal a minus a-r-n over 1 minus r. For which also the numerator, everything's divisible by a, so you can factor out the a, in which case you get 1 minus r to the n over 1 minus r. This gives us the general formula for the sum of terms in a geometric sequence. This gives us a formula for what we call a geometric sum, which we see highlighted right here. s-n is equal to a-1, the first term, which we sometimes just call that 1a, and then we get 1 minus r to the n over 1 minus r. Let's look at an example of this. Let's find the sum, s-n, the geometric sum of the first n terms of a geometric sequence, 1 over 2 to the n, where our sequence looks like 1 half, 1 fourth, 1 eighth. We'll just take powers of 1 half. Let's find the general formula. In this geometric sequence, notice that the first term is 1 half. We're going to plug that in for our a-1 right there. But also, if we start taking consecutive terms, 1 fourth divided by 1 half, that equals 1 half. That's the same thing as 1 eighth divided by 1 fourth, et cetera. This is our r value. We're going to plug in r is likewise 1 half. In which case then we get something like the following. We're going to get 1 half times 1 minus 1 half to the n over 1 minus 1 half. Now, 1 minus 1 half is itself equal to 1 half, and these 1 halves will cancel. You're times e by 1 half and dividing by 1 half. Your final result, the general formula for s-n right here is going to be 1 minus 1 over 2 to the n. We can use this formula to add up together the terms 1 half plus 1 fourth plus 1 eighth, 1 sixteenth, et cetera. We can do this quite quickly in fact. If we want to do s-8, the sum of the first 8 powers of 1 half, so that's 1 half, 1 fourth, 1 eighth, 1 sixteenth, all the way up to 1 over 256, this thing will add up just to be 1 minus 1 over 256. That is 255 over 256. So the numerator will just be 1 less than the denominator, which is the greatest power of 2 you see right there. So we could very quickly find out geometric sums if we know the initial value and if we know the constant ratio that defines our geometric sequence.