Generally Geometric
Solution for internal use NOT for publication
We know that $$\sum_{n=0}^\infty x^n = {1\over(1-x)}$$ so
$$\sum_{n=0}^\infty nx^n = x{\rm{d}\over \rm{dx}}\sum_{n=0}^\infty
x^n $$ $$= x{\rm{d}\over \rm{dx}}\left({1\over 1-x}\right) $$ $$ =
{x\over(1-x)^2} .$$ Going one step further $$\sum_{n=0}^\infty
n^2x^n = x{\rm{d}\over \rm{dx}}{x\over(1-x)^2}$$ $$ =
x\left({(1-x)^2+2x(1-x)\over (1-x)^4}\right) $$
$$=x\left({(1-x)+2x\over (1-x)^3}\right)$$ $$= {x(1+x)\over
(1-x)^3} .$$
More generally $$\sum_{n=0}^\infty n^kx^n = \left(x{\rm{d}\over
\rm{dx}}\right) \cdots \left(x {\rm{d}\over \rm{dx}}\right) \left(
{1\over 1-x}\right) $$ where the operator. $$ \left(x {\rm{d}\over
\rm{dx}}\right)$$ is applied $k$ times.