Well on the left hand side, all I have is a P times the expected value of X. If I want to solve for the expected value of X, I just divide both sides by P. So I get, and this is kind of neat, through this mathematical gymnastics I now have, I'm just dividing everything by P, both sides. On the left hand side, I just have the expected value of X. 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?