 When you add together the term of a sequence, what you get is a series. And so I can start with a geometric series, which again looks like the form A, A times something, times something again, and so on, as far as I care to go. And then I have the corresponding geometric series that consists of the sum of all of our terms. Now here it's very important to remember this dot, dot, dot says there's a whole bunch of things very similar to what we have that we're not going to write down because their form should be pretty obvious. Now I can find the value of this sum, obviously because it's a finite sum of terms, I could just add these things together. On the other hand, it might be nice to have an easier way to find it. Adding together one or two or three terms might not be too bad, but if there's 50 or 100 or a billion of these terms, it's a little bit less convenient to add them two by two. So let's see if we can derive a formula for the sum of a geometric series. And so here's our basic problem. I have a geometric series and I want to find what the sum of that series is going to be. And we'll do something that's very common in mathematics. I don't know what the sum is, but I can still give it a name. So what I'm going to do is I'm going to say this sum is, I don't know what it is, but I'll call it something how about S in a fit of creativity. Now here's the nice thing about it because everything inside is an integer and also everything inside is finite. Then we can do algebra with it. Now note one thing that I've done. Remember this dot dot dot is there's a whole bunch of their terms just like these other ones that we're not bothering to write down. For purposes that'll become apparent in a second, I've written down our starting term, our last term, and I've included at least one of these similar terms. This is a r to the n minus one. So notice the exponent on r goes up by one every time. If n is the last power on r, the one before it will be n minus one. And I've written that for reasons that'll become apparent in a second. Now the crucial observation here is that every term of the series is r times the preceding term. So if I find r times s, well that's r times all of these terms in the series, and what that's going to do is it's going to take each of these terms and produce something that looks very similar to each of these terms. So I'll do a little bit of algebra. rs is ar, ar squared, ar cubed, and so on. And this term that we included here that was there but we didn't bother to write the first time becomes ar to the n. This last term, ar to the n, when I multiply by r, becomes ar to power n plus one. And at this point you say, well I want to find the sum of these n plus one terms. And what you've just done is you've given me not just n plus one terms here, but you've actually doubled the size of the problem to find the sum. There's too many things here. Well, what we would like to do is to get rid of sum of the things that we have to add. And what better way is there to get rid of things than to subtract. So if I take this and I subtract this, well a lot of my terms are going to drop out. So I'll subtract s minus r times s. So that's this ar up here, but I'm subtracting ar down here so that pair goes away. ar squared up here minus ar squared there, that goes away. There's an ar to the third here that I'm not writing down, but then it'll be subtracted, ar to the third here, and so on. This is going to go away, ar to the n minus ar to the n, that goes away. And then at the very end I'm subtracting ar to the n plus one. When I do this subtraction, everything except for the first term of the first sum and the last term of the second sum goes away, and I have a much easier expression to deal with. And at this point I could use a little bit of algebra to solve for this unknown value s. So let's see if in doubt factor out or multiply. As the case may be, there's an s in both of these terms, so I could factor that out. That's s times one minus r still equals left hand side, right hand side, sorry. And let's see if I want to solve for s. I want to divide by one minus r, and there's my nice summation formula for a geometric series. Now, having gone through all this trouble to find the summation formula for the geometric series, here's a little bit of advice. Don't bother memorizing this. Understand this process. If you try to apply the formula nine times out of ten, trying to apply the formula is more of a headache than it's worth. If you understand the process, you can derive the formula every time, and you can find your answer much more reliably than you can by plugging in numbers into a formula. So, for example, let's consider this sum one plus three and so on up to three to the power of fifty. And the first thing we do want to do here is check to make sure that it's a geometric series. The first term is one, and every term appears to be three times the preceding term. This is three times one, this is three times three, and so on all the way up to the end, which is three times a whole bunch of other things. So, again, this is a geometric series, and we could use the geometric series summation formula, but why bother? So, let's go through our derivation again, and again, we don't know what the sum is, but we'll call it s. And again, the reason that the geometric series summation works is that every term of the series is the same number, the common ratio, times the preceding term. Here, every term is three times the preceding term, so I can multiply by the common ratio, three s, and this becomes a one, this becomes a three squared, a three cubed, and so on. This last term, three to the fiftieth, becomes three to the fifty first, and now I'll subtract s minus three s, three minus three, three squared. Everything goes away except first term and last term. Do a little bit of algebra, s minus three s, negative two s, same thing over on the right hand side. Do a little bit of algebra, do a little bit of style, because we don't like having a negative number in the denominator. We'll distribute that, and there is my sum of the series.