Prove that: $\sum\limits_{n=1}^\infty \frac{n^2}{2^n}=6$. Prove that: $$\sum\limits_{n=1}^\infty \dfrac{n^2}{2^n}=6.$$ I am trying to find the sum of this infinite series. Got stuck. Any help will be highly appreciated.
 A: Just for variety, reindexing sums and using the geometric series formula:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{n^2}{2^n}
&=\sum_{n=1}^{\infty}\frac{\sum_{k=1}^{n}(2k-1)}{2^n}\\
&=\sum_{k=1}^{\infty}(2k-1)\sum_{n=k}^{\infty}\frac{1}{2^n}\\
&=\sum_{k=1}^{\infty}(2k-1)\frac{(1/2)^k}{1-1/2}\\
&=2\sum_{k=1}^{\infty}\frac{(2k-1)}{2^k}\\
&=2\sum_{k=1}^{\infty}\frac{k}{2^{k-1}}-2\sum_{k=1}^{\infty}\frac{1}{2^k}\\
&=2\sum_{k=1}^{\infty}\frac{\sum_{j=1}^k1}{2^{k-1}}-2\frac{1/2}{1-1/2}\\
&=2\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}\frac{1}{2^{k-1}}-2\\
&=2\sum_{j=1}^{\infty}\frac{1/2^{j-1}}{1-1/2}-2\\
&=4\sum_{j=1}^{\infty}\frac{1}{2^{j-1}}-2\\
&=4\frac{1}{1-1/2}-2\\
&=6
\end{align}$$
A: prove by induction that $$\sum_{n=1}^k\frac{n^2}{2^n}=2^{-k} \left(-k^2-4 k+3\ 2^{k+1}-6\right)$$
A: Let $S=\sum_{n\geq 1}\frac{n^2}{2^n}$ (the series is obviously convergent). We have:
$$\begin{eqnarray*} S = 2S-S &=& \sum_{n\geq 1}\frac{n^2}{2^{n-1}}-\sum_{n\geq 1}\frac{n^2}{2^n}\\&=&1+\sum_{n\geq 1}\frac{(n+1)^2-n^2}{2^n}\\&=&1+\sum_{n\geq 1}\frac{2n+1}{2^n}=2+2\sum_{n\geq 1}\frac{n}{2^n},\end{eqnarray*}$$
hence it is enough to show that $T=\sum_{n\geq 1}\frac{n}{2^n}=2$. In the same way:
$$ T = 2T-T = 1+\sum_{n\geq 1}\frac{1}{2^n} = 2 $$
and we are done. A similar proof is the following one: for any $x$ with $|x|<1$,
$$ (1-x)^3\sum_{n\geq 1}n^2 x^n =\sum_{n\geq 1}\delta^3(n^2)\,x^n = x+x^2$$
with $\delta$ being the backward difference operator. That gives:
$$ \sum_{n\geq 1}n^2 x^n = \frac{x+x^2}{(1-x)^3}$$
and it is enough to evaluate the previous identity at $x=\frac{1}{2}$.
A: We have,  $$S = \sum_{n=1}^\infty \frac{n^2}{2^n} = \sum_{n=0}^\infty \frac{n^2}{2^n}  $$
$$S = \sum_{n=1}^\infty \frac{n^2}{2^n} = \sum_{n=0}^\infty \frac{(n+1)^2}{2^{n+1}}  $$
Since $S = 2S - S  $,
\begin{align}
S&=\sum_{n=0}^\infty \frac{(n+1)^2}{2^n} - \sum_{n=0}^\infty \frac{n^2}{2^n} \\  
&=\sum_{n=0}^\infty \frac{2n+1}{2^n}\\ &=  
\sum_{n=0}^\infty \frac{2n}{2^n} + \sum_{n=0}^\infty \frac{1}{2^n} \\&=
4+2 = 6
\end{align}
A: Hint: For $|x|<1$:
$\frac{1}{1-x} = \sum_{i=0}^\infty x^i$
$\frac{d}{dx} [\frac{1}{1-x}] = \sum_{i=1}^\infty i x^{i-1}$
$\frac{d^2}{dx^2} [\frac{1}{1-x}] = \sum_{i=2}^\infty i (i-1) x^{i-2}$
Take $x=\frac{1}{2}$ and use these statements with a change of indexing to get $\sum_{i=0}^\infty \frac{i^2}{2^i}$. 
