A calculus problem by Sahil Silare

Using, $\displaystyle \sum_{n=1}^{\infty} n^2 x^n = \frac{x(x+1)}{(1-x)^3} , |x|<1$ & $\displaystyle \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} , |x|<1$ we have ,

$\displaystyle S=\sum_{n=1}^{\infty} \frac{n^2}{2^n} + 2\frac{n}{2^n}+3\frac{1}{2^n} = 6+4+3=13$

Thereferore , $\displaystyle \sum_{r=1}^{s} r^3 = \boxed{8281}$

$\begin{aligned}\\ \displaystyle~&\sum_{n=1}^{\infty}~\dfrac{n^2+2n+3}{ 2^n}\\ \displaystyle~=&\sum_{n=2}^{\infty}~\dfrac{n^2+2} {2^{n-1}}\\ \displaystyle~=&\sum_{n=1}^{\infty}~\dfrac{n^2}{ 2^{n-1}}- \sum_{n=1}^1~\dfrac{n^2}{ 2^{n-1}}+\sum_{n=1}^{\infty}~\dfrac 2 { 2^{n-1}} - \sum_{n=1}^1~\dfrac 2 { 2^{n-1}}\\ =&12-1+4-2=13.\\ \displaystyle~=&\sum_{n=1}^{13}n^3\\ =& \left \{ \dfrac{13(13+1)} 2 \right \}^2\\ =& \Large~~\color{#D61F06}{8281}.\\ \end{aligned}$

$\displaystyle\\ \sum_{n=1}^{\infty} n^2*x^n=\dfrac{x(x+1)}{(1-x)^3}\\ When~x=2^{-1},\\ \sum_{n=1}^{\infty} 2*n^2*x^{n-1}=2*\dfrac{x(x+1)}{(1-x)^3} \\ = 2*\dfrac {\frac1 2*(\frac1 2+1)}{(1-\frac 1 2)^3}\\ =2*\dfrac{\frac1 2*\frac3 2}{\frac 1 8}=12$
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