Sum of a series - 5 Can anyone simplify this series,
$$S= \sum_{n=2}^{\infty} \frac{n(n-1)}{2^{2n-1}}$$ into a fraction?
Is there a closed form solution for this?
This seems to the answer for one other problem that I am working on?
Any help is appreciated.

HINT:
Let $\displaystyle S=\sum_{r=0}^nr(r-1)a^r=\sum_{r=2}^nr(r-1)a^r  $
$\displaystyle\implies a\cdot S=\sum_{r=2}^nr(r-1)a^{r+1}=\sum_{r=3}^{n+1}(r-1)(r-2)a^r $
$\displaystyle\implies S(1-a)=2a^2+\sum_{r=3}^n2(r-1)a^r-n(n-1)a^{n+1}$
$\displaystyle S(1-a)=2a^2-n(n-1)a^{n+1}+2a\sum_{r=3}^n(r-1)a^{r-1}$
Again let $\displaystyle T=\sum_{r=3}^n(r-1)a^{r-1}$
$\displaystyle T-T\cdot a=?$
Can you recognize the Geometric Series?
Now for $|a|<1,$ use Sequence Limit: $\lim\limits_{n \rightarrow \infty}{n\,x^n}$ or How to prove that $\lim_{n \to \infty} n x^{n} = 0 $ when $0<x<1$?
Can you prove $\lim_{n\to\infty}n(n-1)a^n=0$ for $|a|<1?$
