I presume you know that for any $x$, $\sum _{n=0}^{\infty }{x}^n=1/(1-x)$. Try differentiating both sides of this equation: does this help at all?

Define

$S=\sum _{n=1}^{\infty }{n}^2/2^n$

Then

$S-S/2=\sum _{n=1}^{\infty }{n}^2/2^n-n^2/2^{n+1}$

$S-S/2=1/2+\sum _{n=2}^{\infty }[n^2-(n+1{)}^2]/2^n$

$S-S/2=1/2+\sum _{n=2}^{\infty }[2n-1]/2^n$

The latter sum splits into $2\sum _{n=2}^{\infty }n/2^n$ and $\sum _{n=2}^{\infty }1/2^n$. You probably know how to do the latter sum, and the former sum can be written in terms of the latter by the same technique that was used on $S$.