How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know what to do with the $n^2$. 

We will use the same technique found here to show that $\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$. But we'll need the following theory extension (see this link) to that result. For $k \ge 1$,
$\tag 1 \sum_{n=k}^{\infty}\frac{n}{2^n} = 2^{1 - k} \, (k + 1)$
We decompose the summands of $\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ in a natural/straightforward manner (so that $\text{(1)}$ applies) , arranging these numbers into a table:
$$\begin{pmatrix}
\frac{1}{2} & \frac{2}{4}  & \frac{3}{8} & \frac{4}{16} & \frac{5}{32} & \dots \\
0 & \frac{2}{4}  & \frac{3}{8} & \frac{4}{16} & \frac{5}{32} & \dots \\
0 & 0  & \frac{3}{8} & \frac{4}{16} & \frac{5}{32} & \dots \\
0 & 0  & 0 & \frac{4}{16} & \frac{5}{32} & \dots \\
0 & 0 & 0 & 0 & \frac{5}{32} & \dots \\
0 & 0 & 0 & 0 & 0 & \dots \\
. \\
. \\
. \\
\end{pmatrix}$$
Using $\text{(1)}$ we add up the entries in each row, giving the sequence of numbers
$\tag 2 \bigr ( \, 2^{1 - k} \, (k + 1) \, {\bigr )}_{\,k \ge 1}$
Performing the summation,
$\quad  \sum_{k=1}^{\infty} 2^{1 - k} \, (k + 1) =  2 \sum_{k=1}^{\infty} \frac{k + 1}{2^k} = 2 \big ( \sum_{k=1}^{\infty} \frac{k}{2^k} +  \sum_{k=1}^{\infty} \frac{1}{2^k}  \big)$
$\quad \quad  2 \sum_{k=1}^{\infty} \frac{k + 1}{2^k} = 2(2+1) = 6$
So adding up all those numbers in the table gives
$\quad \text{ANS: } 6$
