Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$ Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$
My approach : 
$= \frac{1}{6}+\frac{2^2}{6^2}+\frac{3^2}{6^3} +\cdots \infty$
Now how to solve this I am not getting any clue on this please help thanks.
 A: Starting with 
\begin{align}
\frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n}
\end{align}
then differentiate and multiply by $t$ to obtain
\begin{align}
\frac{t}{(1-t)^{2}} = \sum_{n=0}^{\infty} n t^{n}.
\end{align}
Repeating leads to
\begin{align}
\sum_{n=0}^{\infty} n^{2} \ t^{n} = \frac{t(1+t)}{(1-t)^{3}}.
\end{align}
Now let $t = 1/6$ for which
\begin{align}
\sum_{n=0}^{\infty} \frac{n^{2}}{6^{n}} = \frac{42}{125}.
\end{align}
A: Break $n^2$ into two parts: $\underbrace{n(n-1)} + \underbrace{{}\quad n\quad {}}$.
The first part appears in the second derivative of $x^n$ and the second part in the first derivative:
\begin{align}
\sum_{n=1}^\infty n^2 x^n & = \sum_{n=1}^\infty x^2\Big(n(n-1) x^{n-2} \Big) + x\Big(nx^{n-1}\Big) \\[10pt]
& = \left(x^2 \frac{d^2}{dx^2} \sum_{n=1}^\infty x^n\right) + \left( x\frac{d}{dx}\sum_{n=1}^\infty x^n \right) \\[10pt]
& = \left( x^2 \frac{d^2}{dx^2} \frac{x}{1-x}\right) +  \left( x\frac{d}{dx} \frac{x}{1-x} \right) \\[10pt]
& = \cdots\cdots
\end{align}
(Finally at the end, put $x=1/6$.)
