<s>[INST] Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$
I really tried, but I couldn't, help guys?
 [/INST] Since $n^2 = 2\binom{n}{2}+\binom{n}{1}$ we have:
$$ S=\sum_{n\geq 1}\frac{n^2}{2^n}=\left.\left(2\frac{x^2}{(1-x)^3}+\frac{x}{(1-x)^2}\right)\right|_{x=\frac{1}{2}}=\color{red}{6}.$$
As an alternative to the negative binomial series, we may also use:
$$ S = 2S-S = \sum_{n\geq 1}\frac{n^2}{2^{n-1}}-\sum_{n\geq 1}\frac{n^2}{2^n}=1+\sum_{n\geq 1}\frac{(n+1)^2-n^2}{2^n}=2+2\sum_{n\geq 1}\frac{n}{2^n}$$
so that, in the same way:
$$ \sum_{n\geq 1}\frac{n}{2^n} = T = 2T-T = \sum_{n\geq 1}\frac{n}{2^{n-1}}-\sum_{n\geq 1}\frac{n}{2^n} = 1+\sum_{n\geq 1}\frac{1}{2^n}=2 $$
and $S=2+2\cdot 2=\color{red}{6}$.
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