If I divide all of these terms by P, this first term becomes one, the second term becomes one minus P. This third term, if I divide by P, becomes plus one minus P squared, so forth and so on. Now what's cool about this, this is a classic geometric series with a common ratio of one minus P. And if that term is completely unfamiliar to you, I encourage you, and this is why it's actually called a geometric, one of the reasons, arguments for why it's called a geometric random variable, but I encourage you to review what a geometric series is on Khan Academy if this looks completely unfamiliar. But in other places, we prove, using actually a very similar technique that we did up here, that this sum is going to be equal to one over one minus our common ratio. And our common ratio is one minus P. So what is this going to be equal to? And we are really in the home stretch right over here. This is going to be equal to one over one minus one plus P. One minus one plus P, which is indeed equal to one over P. So there you have it. We have proven to ourselves that the expected value of a geometric random variable, using some, I think, cool mathematics, is indeed equal to one over P.