Question

The value of $\sum _{k=0}^{\infty }\frac{k^2}{4^k}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Then the value of $(m+n)$ is

• A. $28$
• B. $47$
• C. $66$
• D. $85$

Answer 1

The correct option is B $47$

Given $S=\sum _{K=0}^{\infty }\frac{k^2}{4^k}=\frac{1^2}{4}+\frac{2^2}{4^2}+\frac{3^2}{4^3}+\frac{4^2}{4}+...+\frac{k^2}{4k}+...\left(1\right)$

$\frac{1}{4}S=\frac{1}{4}\sum _{K=0}^{\infty }\frac{k^2}{4^k}=\frac{1^2}{4^2}+\frac{2^2}{4^3}+\frac{3^2}{4^4}+...+\frac{(k-1{)}^2}{4^k}+...\left(2\right)$

Subtracting (2) from (1)

$\Rightarrow \frac{3S}{4}=\frac{1}{4}+\frac{3}{4^2}+\frac{5}{4^3}+\frac{7}{4^4}+...\left(3\right)$ (A.G.P series)

$\frac{3S}{16}=\frac{1}{4^2}+\frac{3}{4^3}+\frac{5}{4^4}+...\left(4\right)$

$\frac{9S}{16}=\frac{1}{4}+\frac{2}{4^2}+\frac{2}{4^3}+\frac{2}{4^4}...$

$=\frac{1}{4}+\frac{2}{4^2}(1+\frac{1}{4}+\frac{1}{4^2}+...)$

$=\frac{1}{4}+\frac{2}{16}\times \frac{1}{1-\frac{1}{4}}$ [infinte G.P sum $\frac{a}{1-r}]$

$=\frac{1}{4}+\frac{1}{8}\times \frac{4}{3}=\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

$S=\frac{5}{12}\times \frac{16}{9}=\frac{20}{27}$

$m=20,n=27\Rightarrow (m+n)=47$

$\therefore$ option B is correct.