<s>[INST] 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.
 [/INST] 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}
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