How to compute the sum $\sum_{n=0}^\infty \tfrac{n^2}{2^n}$? 
How to find this sum :
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}$

$\sum_{n=0}^\infty \dfrac{n^2}{2^n}=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{9}{8}+\dfrac{16}{16}+\dfrac{25}{32}+\dfrac{36}{64}+\dfrac{49}{128}+\dots$
Now $\sum_{n=0}^\infty \dfrac{n}{2^n}\leqslant \sum_{n=0}^\infty \dfrac{n^2}{2^n}$
And I know that $\sum_{n=0}^\infty \dfrac{n}{2^n}=2$.
But how to find this sum ? I am confused.Please give some hints.

Interesting... let's look at it.
$S=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+...$
$\frac{S}{2}=\frac{1}{4}+\frac{4}{8}+\frac{9}{16}+...$
$\frac{S}{2}=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+...$
$\frac{S}{4}=\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+...$
$\frac{S}{4}=\frac{1}{2}+\frac{2}{4}+\frac{2}{8}+...=\frac{1}{2}+1=\frac{3}{2}$
So $S=6.$
EDIT: According to the comments this is wrong. I'll check over it (?)
EDIT2: Apparently that solution (which is for $\Sigma_{n=0}^{\infty}\frac{n}{2^n}$) gives 2 for the original summand. I can't read, oops. :\
