trouble calculating sum of the series $ \sum\left(\frac{n^2}{2^n}\right) $

Find out the sum of the series $\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{n^2}{ 2^n}$. I have checked the convergence, but how to calculate the sum?

Note $\sum\limits_{n\geqslant 0} t^n = \dfrac{1}{ 1-t}$ gives $\sum\limits_{n\geqslant 1} nt^n = \dfrac{t}{( 1 - t)^2}$ after differentiation and multiplication by $t$, which in turn gives $\sum\limits_{n\geqslant 1} n^2t^n = \dfrac{t(t+1)}{( 1 - t)^3}$ by the same token.

Begin with a geometric series, $$\sum_{n=0}^\infty z^n = {1\over 1 - z}.$$ if $|z| < 1$. Differentiate and you get $$\sum_{n=1}^\infty nz^{n-1} = {1\over (1-z)^2}.$$ Do it again $$\sum_{n=2}^\infty n(n-1)z^{n-2} = {1\over (1-z)^3}.$$ Massage these, plug in $z = 1/2$ and you will get it.