Show that $\sum_{k=0}^{\infty}k^2x^k= \frac{x(1+x)}{(1-x)^3}$ for $0 < |x| < 1$ Show that $$\sum_{k=0}^{\infty}k^2x^k= \frac{x(1+x)}{(1-x)^3}$$ for $0 < |x| < 1$.
(This is appendix question A.1-3 from Introduction to Algorithms by Cormen)
 A: Hint: $$\sum_{k=0}^{\infty}k^2x^k= x\sum_{k=0}^{\infty}k^2x^{k-1}= x\frac{d}{dx}\sum_{k=0}^{\infty}kx^k$$
Can you follow on from this?
Edit: some more detail:
The idea with a question like this is to transform the question into something we already know. In this case, the sum looks remarkably similar to $$\sum_{k=0}^\infty x^k = \frac{1}{1-x} \text{ for } 0<x<1$$
Can you see a way to transform (perhaps by differentiating) the original sum into the simpler one above?
Continuation (if you get stuck):
$$x\frac{d}{dx}\sum_{k=0}^{\infty}kx^k = x\frac{d}{dx}\left(  x\sum_{k=0}^{\infty}kx^{k-1}\right)= x\frac{d}{dx}\left(  x\frac{d}{dx}\sum_{k=0}^{\infty}x^k\right)= x\frac{d}{dx}\left(  x\frac{d}{dx}\left(\frac{1}{1-x}\right)\right)$$
A: To find the required sum you could just find the sum of the geometric progression $1,x,x^2,x^3,... \space where \space |x|\lt 1,$ and then double differentiate it .$$\sum_{k=0}^\infty x^k=\frac 1{1-x}\\=>\frac d{dx}\left(\sum_{k=0}^\infty x^k\right)=\frac d{dx}\left(\frac 1{1-x}\right)\\=>\sum_{k=0}^\infty kx^{k-1}=\frac 1{(1-x)^2}\\=>\sum_{k=0}^\infty kx^k=\frac x{(1-x)^2}\\=>\frac d{dx}\left(\sum_{k=0}^\infty kx^k\right)=\frac d{dx}\left(\frac x{(1-x)^2}\right)\\=>\sum_{k=0}^\infty k^2x^{k-1}=\frac{(1-x)^2+2(1-x)x}{(1-x)^4}\\=>\sum_{k=0}^\infty k^2x^{k-1}=\frac{1+x}{(1-x)^3}\\=>\sum_{k=0}^\infty k^2x^k=\frac {x(1+x)}{(1-x)^3}$$
