Evaluating $\sum_{n=0}^\infty n^2 x^n $ by splitting it into two summations?

Question

How to find the sum of the following summation?

$$\sum_{n=0}^\infty n^2 x^n $$

I was told to split it into two summations, but I don't see any help in that.

(original problem image)

Answer 1

A few people have given elements of a solution in the comments, but for the sake of clarity I'll write it out here. As @Eevee Trainer said, we have the equality $$\sum_{n=0}^\infty x^n=\frac 1{1-x}.$$ As @runway44 suggested, differentiating this equation once gives $$\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}.$$ Differentiating it again, we get $$\sum_{n=0}^\infty n(n-1)x^{n-2}=\frac{2}{(1-x)^3}.$$ This is why I suggested the split $$\sum_{n=0}^\infty n^2x^n=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+x\sum_{n=0}^\infty nx^{n-1}.$$ Using the previous calculations, we conclude that $$\sum_{n=0}^\infty n^2x^n=\frac{2x^2}{(1-x)^3}+\frac{x}{(1-x)^2}=\frac{x^2+x}{(1-x)^3}.$$