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.

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}
