Calculating $\sum_{n=0}^\infty\frac{(n+1)(n+2)}{2}x^n$ when $|x|<1$

Question

How do I calculate the sum $$\sum_{n=0}^\infty\frac{(n+1)(n+2)}{2}x^n$$ I know this sum will be finite for $|x|<1$ and will diverge for other values of $x$. Since for other sums it was common to derivate in order to have a sum in the form $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ I thought it would be a good idea to integrate. However I sill can't solve it. Wolfram Alpha says the sum is $\frac{-1}{(x-1)^3}$. I would appreciate if someone could guide me to this result.

Answer 1

The hint:

Calculate $$\frac{1}{2}\left(\sum_{n=0}^{+\infty}x^{n+2}\right)''.$$

Answer 2

Consider $f(x)=\sum_{n\geq 0}{x^{n+2}}$ and differentiate twice. Don’t forget to justify the differentiation under summation.