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.
 A: 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. :\
A: 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. 
A: Hint: For $|x|<1$
\begin{align*}
\sum_{n=0}^{\infty }n^{2}x^{n}&=\sum_{n=0}^{\infty }n\left ( n+1 \right )x^{n}-\sum_{n=0}^{\infty }nx^{n} \\
&=x\sum_{n=0}^{\infty }n\left ( n+1 \right )x^{n-1}-x\sum_{n=0}^{\infty }nx^{n-1}\\
&=x\left ( \sum_{n=0}^{\infty }x^{n+1} \right )''-x\left ( \sum_{n=0}^{\infty }x^{n} \right )'\\
&=-\frac{x\left ( x+1 \right )}{\left ( x-1 \right )^{3}}
\end{align*}
then let $x=\dfrac{1}{2}$ you will get the answer.
