Evaluating $\sum\limits_{x=1}^\infty x^2\cdot\left(\frac{1}{2}\right)^{x+1}$? How can I calculate $\sum\limits_{x=1}^\infty x^2\cdot\left(\frac{1}{2}\right)^{x+1}$?
It's a follow-up question to $\sum\limits_{x=1}^\infty x\cdot\left(\frac{1}{2}\right)^{x+1}$, which I found by factoring out $\left(\frac{1}{2}\right)^2$ and then looking at the sum as a derivative of a geometric series, but the same approach does not work here. I've tried manipulating $x^2$ into a form similar to $(x+1)(x-1)+1$ but it hasn't gotten me very far.
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With $\ds{\verts{\mu} < 1}$:
\begin{align}
\sum_{x = 1}^{\infty}\mu^{x} &= {\mu \over 1 - \mu} = -1 + {1 \over 1 - \mu}
\end{align}

Derivate respect $\ds{\mu}$:
  \begin{align}
\sum_{x = 1}^{\infty}x\mu^{x - 1} &={1 \over \pars{1 - \mu}^{2}}
\quad\imp\quad\sum_{x = 1}^{\infty}x\mu^{x} ={\mu \over \pars{1 - \mu}^{2}}
\end{align}

Derivate again:
\begin{align}
\sum_{x = 1}^{\infty}x^{2}\mu^{x - 1} &=-\,{1 + \mu\over \pars{1 - \mu}^{3}}
\quad\imp\quad
\sum_{x = 1}^{\infty}x^{2}\mu^{x + 1}=
-\,{\mu^{2}\pars{1 + \mu}\over \pars{1 - \mu}^{3}}
\end{align}

Replace $\ds{\mu = \half}$:
  $$
\color{#00f}{\large\sum_{x = 1}^{\infty}x^{2}\pars{\half}^{x + 1}}=
\left.-\,{\mu^{2}\pars{1 + \mu}\over \pars{1 - \mu}^{3}}\right\vert_{\mu\ =\ 1/2}
=\color{#00f}{\large 3}
$$

A: One of the ways is perturbation method from 'Concrete Mathematics'. Consider
$$
S_{n} = \sum_{k=1}^{n} k^3 a^{k+1}\\
S_{n+1} = S_n + (n+1)^3 a^{n+2} = \sum_{k=1}^{n+1}k^3 a^{k+1} = \sum_{k=1}^{n}(k+1)^3 a^{k+1} \\
=S_n +3 \sum_{k=1}^{n}k^2 a ^{k+1} +3 \sum_{k=1}^nka^{k+1} + \sum_{k=1}^{n}a^{k+1} 
$$
Obviously $S_n$ cancels out, you already know the expression for $\sum_{k=1}^{n} k a^{k+1}$, so you can do the rest. Be careful with the algebra though. 
