How to find the sum of the following series How can I find the sum of the following series?
$$
\sum_{n=0}^{+\infty}\frac{n^2}{2^n}
$$
I know that it converges, and Wolfram Alpha tells me that its sum is 6 .
Which technique should I use to prove that the sum is 6?

It is equal to $f(x)=\sum_{n \geq 0} n^2 x^n$ evaluated at $x=1/2$.
To compute this function of $x$, write $n^2 = (n+1)(n+2)-3(n+1)+1$, so that $f(x)=a(x)+b(x)+c(x)$ with:
$a(x)= \sum_{n \geq 0} (n+1)(n+2) x^n = \frac{d^2}{dx^2} \left( \sum_{n \geq 0} x^n\right) = \frac{2}{(1-x)^3}$
$b(x)=\sum_{n \geq 0} 3 (n+1) x^n = 3\frac{d}{dx} \left( \sum_{n \geq 0} x^n \right) = \frac{3}{(1-x)^2}$
$c(x)= \sum_{n \geq 0} x^n = \frac{1}{1-x}$
So $f(1/2)=\frac{2}{(1/2)^3}-\frac{3}{(1/2)^2} + \frac{1}{1/2} = 16-12+2=6$.
The "technique" is to add a parameter in the series, to make the multiplication by $n+1$ appear as differentiation.
