Progression (Sequence and Series)HardBloom L3

Question

The sum of infinite series

$1^{2} - \frac{2^{2}}{5} + \frac{3^{2}}{5^{2}} - \frac{4^{2}}{5^{3}} + \frac{5^{2}}{5^{4}} - \frac{6^{2}}{5^{2}} + \ldots\ldots\ldots\infty$ is equal to

Options

A.$\frac{5}{27}$
B.$\frac{25}{27}$
C.$\frac{25}{108}$
D.$\frac{25}{54}$

Solution

Sol. [Key Concept: Arithmetic–geometric series: use $S - rS$ to collapse it.]

$$\begin{matrix} S & \ = \sum_{r = 1}^{\infty}\mspace{2mu}\mspace{2mu} r^{2}\left( - \frac{1}{5} \right)^{r - 1} \\ \sum_{r = 0}^{\infty}\mspace{2mu}\mspace{2mu} x^{r} & \ = \frac{1}{1 - x} \\ \Rightarrow \ \sum_{r = 1}^{\infty}\mspace{2mu}\mspace{2mu} rx^{r} = \frac{x}{(1 - x)^{2}} & \ = \frac{1}{x - 1} + \frac{1}{(x - 1)^{2}} \\ \Rightarrow \ \sum_{r = 1}^{\infty}\mspace{2mu}\mspace{2mu} r^{2}x^{r - 1} & \ = - \frac{1}{(x - 1)^{2}} - \frac{2}{(x - 1)^{3}} \\ \Rightarrow \ \sum_{r = 1}^{\infty}\mspace{2mu}\mspace{2mu} r^{2}\left( - \frac{1}{5} \right)^{r - 1} & \ = \frac{25}{54} \end{matrix}$$

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