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.

If $-1< x < 1$, we have, by differentiation and adding :
$$\begin{align} \sum_{n=0}^\infty x^n & = \frac 1{1-x} \implies &  \\ \sum_{n=1}^\infty nx^{n-1} & = \frac 1{(1-x)^2} \implies \\ \sum_{n=1}^\infty nx^n & = \frac x{(1-x)^2}  \implies & \small \\ \sum_{n=1}^\infty n^2x^{n-1} & = \frac {1+x}{(1-x)^3} \implies & \small \\ \sum_{n=1}^\infty n^2x^n & = \frac {x(1+x)}{(1-x)^3} \end{align}$$
Put $x=\frac{1}{2}$. We are done. 
