Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$
I really tried, but I couldn't, help guys?

Here's a bit of a twist on the second approach in Jack D'Aurizio's answer.  It's main virtue (if any) is that it avoids explicitly evaluating the auxiliary sum $\sum{n\over2^n}$, getting it to drop out instead.
It's convenient to start the sum at $n=0$ instead of $n=1$.  Borrowing Jack's notation, we have
$$S=\sum_{n=0}^\infty{n^2\over2^n},\quad T=\sum_{n=0}^\infty{n\over2^n},\quad\text{and}\quad2=\sum_{n=0}^\infty{1\over2^n}$$
Then
$$\begin{align}
S-2&=\sum_{n=0}^\infty{n^2-1\over2^n}\\
&=\sum_{n=0}^\infty{(n+1)(n-1)\over2^n}\\
&=\sum_{m=1}^\infty{m(m-2)\over2^{m-1}}\\
&=2\sum_{m=1}^\infty{m^2\over2^m}-4\sum_{m=1}^\infty{m\over2^m}\\
&=2S-4T
\end{align}$$
so
$$S=4T-2$$
which will turn out to be better written as
$$4S=16T-8$$
Likewise (and here's where the twist comes in), we have
$$\begin{align}
S-8&=\sum_{n=0}^\infty{n^2-4\over2^n}\\
&=\sum_{n=0}^\infty{(n+2)(n-2)\over2^n}\\
&=\sum_{m=2}^\infty{m(m-4)\over2^{m-2}}\\
&=4\sum_{m=2}^\infty{m^2\over2^m}-16\sum_{m=2}^\infty{m\over2^m}\\
&=4(S-{1\over2})-16(T-{1\over2})\\
&=4S-16T+6
\end{align}$$
so
$$3S=16T-14$$
Subtracting this equation from the cleverly written $4S=16T-8$ leaves
$$S=6$$
As promised, we haven't bothered computing $T$ (although it's clear its value is easily obtained).
