# Finding the Sum of a particular Power Series

1. Nov 2, 2004

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.

2. Nov 2, 2004

### Tide

If

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

then

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

3. Nov 2, 2004