Partial sums: formula for nth term from partial sum | Series | AP Calculus BC | Khan Academy

Partial sum of the series we're going from one to infinity summing it up of a sub n is given by, and they tell us the formula for the sum of the first n terms. They say write a rule for what the actual nth term is going to be.

Now to help us with this, let me just create a little visualization here. So if I have a sub 1 plus a sub 2 plus a sub 3, and I keep adding all the way to a sub n minus one plus a sub n, this whole thing, this whole thing that I just wrote out, that is sub sub n. This whole thing is s sub n, which is equal to n + 1 over n + 10.

Now, if I want to figure out a sub n, which is the goal of this exercise, well, I could subtract out the sum of the first n minus one terms. So I could subtract out this, so that is s sub n minus one. And what would that be equal to? Well, wherever we see an n, we'd replace with an n minus one, so it would be n plus 1 over n - 1 + 10, which is equal to n over n + 9.

So if you subtract the red stuff from the blue stuff, all you're going to be left with is the thing that we want to solve for. You're going to be left with a sub n. So we could write down a sub n is equal to s sub n minus s sub n minus one. Or we could write that as equal to this stuff.

So this is n + 1 over n + 10 minus n over n + 9. And this by itself, this is a rule for a sub n. But we could combine these terms, add these two fractions together, and this is actually going to be the case for n greater than one. For n equals 1, s sub one is going to be, well, you can just say a sub one is going to be equal to s sub one.

But then for any other n, we could use this right over here. And if we want to simplify this, well, we can add these two fractions. We can add these two fractions by having a common denominator. So let's see, if we multiply the numerator and denominator here by n + 9, we are going to get so this is equal to n + 1 * n + 9 over n + 10 * n + 9.

And from that, we are going to subtract, let's multiply the numerator and the denominator here by n + 10. So we have n * n + 10 over n + 9 * n + 10.

n + 9 * n + 10, and what does that give us? So let's see, if we simplify up here, we're going to have this is n^2 + 10n + 9, that's that. And then this right over here is n^2 + oh, this is n^2 + 10n, doing that red color, so this is n^2 + 10n.

And remember we're going to subtract this, and so, and we are close to deserving a drum roll. A sub n is going to be equal to our denominator right over here is n + 9 * n + 10, and we're going to subtract the red stuff from the blue stuff.

So you subtract an n from an n squared, those cancel out. Subtract a 10n from a 10n, those cancel out, and you're just left with that blue nine. So there you have it, we've expressed, we've written a rule for a sub n for n greater than one.