how to find the sum for $\sum_{k=2}^{\infty} k(k-1)x^{k-2}$?

Question

I know that the convergence radius is $R = (-1,1)$, but I don't know how to sum this series, please help.

Answer 1

Differentiate $\sum_{k\ge2}x^k=\frac{x^2}{1-x}=-1-x+\frac{1}{1-x}$ twice to get$$\sum_{k\ge2}k(k-1)x^{k-2}=\frac{d^2}{dx^2}\left(-1-x+\frac{1}{1-x}\right)=\frac{d^2}{dx^2}\frac{1}{1-x}=\frac{2}{(1-x)^3}.$$As @metamorphy notes, if we change the lower limit to $k=0$, the second derivative doesn't change because $\frac{d^2}{dx^2}(1+x)=0$, but the revised sum is easily written as $\frac{1}{1-x}$.