Sum: $\sum_{k=1}^{\infty}{2k^2}/{5^{k}}$

Question

$$\sum_{k=1}^{\infty}\frac{2k^2}{5^k}$$

I came across this problem and I didn't get how to solve it at all. May someone explain this to me?

Answer 1

Hint: Write your sum as $$\sum_{i=1}^{\infty} 2k^2x^k $$ with $x=1/5$. That's a power series, so you only need to find which functions has has coefficients $2k^2$ in its power series.

(Further hint: Start with the well-known power series for $\frac{1}{1-x}$. Differentiate it once, multiply by $x$ and differentiate it once again).

Answer 2

Let $S=\sum_{k\geq 1}\frac{2k^2}{5^k}$. Then $$\begin{eqnarray*} 4S=5S-S=\sum_{k\geq 1}\frac{2k^2}{5^{k-1}}-\sum_{k\geq 1}\frac{2k^2}{5^k}&=&\sum_{k\geq 0}\frac{2(k+1)^2}{5^k}-\sum_{k\geq 1}\frac{2k^2}{5^k}\\&=&2+\sum_{k\geq 1}\frac{2(2k+1)}{5^k}\end{eqnarray*}$$ hence $S=\frac{1}{2}+\frac{1}{2}\sum_{k\geq 1}\frac{2k+1}{5^k}=\frac{1}{2}+\frac{T}{2}$. Now you may apply the same trick to $T$: $$ 4T=5T-T = \sum_{k\geq 0}\frac{2k+3}{5^k}-\sum_{k\geq 1}\frac{2k+1}{5^k} = 3+\sum_{k\geq 1}\frac{2}{5^k} = 3+\frac{\frac{2}{5}}{1-\frac{1}{5}}=\frac{7}{2} $$ hence $T=\frac{7}{8}$ and $S=\frac{15}{16}$.