Finding the Sum of a particular Power Series

[DivGradCurl]

I need to obtain the sum of the following series

\sum _{n=1} ^{\infty} \frac{n^2}{2^n}

Well, with the aid of Mathematica, I get the answer, which is 6. What I'm trying to do now is work my way backwards from there. I need to express it through the geometric series

\sum _{n=0} ^{\infty} x^n = \frac{1}{1-x}

In fact, my guess is that I should use

\sum _{n=1} ^{\infty} n x^{n-1} = \frac{1}{\left( x - 1 \right) ^2}

but it can't get it to fit in.

Thank you very much.

 

[Tide]

If

S = \sum_{n=0}^{\infty} x^n

then

\sum_{n=0}^{\infty}n^2 x^n = x \frac {d}{dx} x \frac {dS}{dx}

 

[DivGradCurl]

Thank you so much. This definitely works out!