<s>[INST] Calculation of an infinite sum $\sum_{k=1}^{\infty} \frac{k^3}{2^k} = 26$ Could you give me some advice on how to calculate the following sum?
$$\begin{aligned}\sum_{k=1}^{\infty} \frac{k^3}{2^k} = 26\end{aligned}$$
Thank you!
 [/INST] $$
\begin{aligned}
& S_{0} = \sum_{n=1}^{\infty} \frac{n^{0}}{2^{n}} = \sum_{n=1}^{\infty} \frac{1}{2^{n}} \Rightarrow \color{red}{S_{0} = 1} \\ \\
& S_{1} = \sum_{n=1}^{\infty} \frac{n^{1}}{2^{n}} = \sum_{n=1}^{\infty} \frac{n + 1 - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{n + 1}{2^{n}} - S_{0} \\
& \qquad \Rightarrow S_{1} + S_{0} = \sum_{n=1}^{\infty} \frac{n + 1}{2^{n}} = 2 \sum_{n=1}^{\infty} \frac{n + 1}{2^{n + 1}} = 2 \sum_{n=2}^{\infty} \frac{n}{2^{n}} = 2 \left[ - \frac{1}{2} + \sum_{n=1}^{\infty} \frac{n}{2^{n}} \right] = - 1 + 2 S_{1} \\
& \qquad \Rightarrow \color{red}{S_{1} = 1 (S_{0} + 1) = 2} \\ \\
& S_{2} = \sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}} = \sum_{n=1}^{\infty} \frac{n^{2} + 2 n + 1 - 2 n - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n}} - 2 S_{1} - S_{0} \\
& \qquad \Rightarrow S_{2} + 2 S_{1} + S_{0} = S_{2} + 3 (S_{0} + 1) - 1 =  2 \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n + 1}} = - 1 + 2 S_{2} \\
& \qquad \Rightarrow \color{red}{S_{2} = 3 (S_{0} + 1) = 6} \\ \\
& S_{3} = \sum_{n=1}^{\infty} \frac{n^{3}}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{3} - 3 n^{2} - 3 n - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{3}}{2^{n}} - 3 S_{2} - 3 S_{1} - S_{0} \\
& \qquad \Rightarrow S_{3} + 3 S_{2} + 3 S_{1} + S_{0} = S_{3} + 13 (S_{0} + 1) - 1 =  2 \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n + 1}} = - 1 + 2 S_{3} \\
& \qquad \Rightarrow \color{red}{S_{3} = 13 (S_{0} + 1) = 26} \\ \\
& \text{... etc}
\end{aligned}
$$
</s>