A calculus problem by Raghu Raman Ravi

$\begin{aligned} S_1 & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n}{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {n}{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {n+1}{2^{n+1}} = \frac 12 \left(\sum_{n=0}^\infty \frac {n}{2^n} + \sum_{n=0}^\infty \frac 1{2^n} \right) = \frac 12 S_1 + 1 = 2 \end{aligned}$

$\begin{aligned} S_2 & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n^2}{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {n^2}{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {(n+1)^2}{2^{n+1}} = \frac 12 \sum_{n=0}^\infty \frac {n^2+2n+1}{2^n} = \frac 12 S_2 + S_1 + 1 = 6 \end{aligned}$

$\begin{aligned} S_3 & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n^3}{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {n^3}{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {(n+1)^3}{2^{n+1}} = \frac 12 \sum_{n=0}^\infty \frac {n^3+3n^2+3n+1}{2^n} \\ & = \frac 12 S_3 + \frac 32 S_2 + \frac 32 S_1 + 1 = 2 (9+3+1) = \boxed{26} \end{aligned}$

Here's how I did it. Start with the formula $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ . Differentiate it to obtain $\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$ . Differentiate it again to obtain $\sum_{n=2}^\infty n(n-1)x^{n-2}=\sum_{n=1}^\infty n(n+1)x^{n-1}=\frac{2}{(1-x)^3}$ . Finally differentiate a third time to obtain $\sum_{n=2}^\infty n(n+1)(n-1)x^{n-2}=\sum_{n=1}^\infty n(n+1)(n+2)x^{n-1}=\frac{6}{(1-x)^4}$ . Substitute $x=\frac{1}{2}$ into each of the equations, and then multiply both sides of each equation by $\frac{1}{2}$ to obtain: $\sum_{n=1}^\infty \frac{n}{2^n}=2,\sum_{n=1}^\infty \frac{n^2+n}{2^n}=8,\sum_{n=1}^\infty \frac{n^3+3n^2+2n}{2^n}=48$ . Then, taking a certain linear combination of the equations will lead to the result. So I found $a,b,c$ such that $an+b(n^2+n)+c(n^3+3n^2+2n)=n^3$ . This amounts to solving the system $a+b+2c=0,b+3c=0,c=1$ , which is easily solved using back substitution as $a=1,b=-3,c=1$ . So the answer is $2a+8b+48c=2-24+48=26$ .