Question

If sum of the series $1+\frac{1}{5}+\frac{3}{5^2}+\frac{5}{5^3}+\dots \infty$ is $k$. Find $8k$

Answer 1

Let $k=1+\frac{1}{5}+\frac{3}{5^2}+\frac{5}{5^3}+\dots \infty$ ...... (i)

$\frac{k}{5}=\frac{1}{5}+\frac{1}{5^2}+\frac{3}{5^3}+\frac{5}{5^4}+\dots \infty$ .......(ii)

Now subtracting (ii) from (i) by shifting one place, we get

$\frac{4k}{5}=1+\frac{2}{5^2}+\frac{2}{5^3}+\frac{2}{5^4}+\dots \infty$

$=1+\frac{2}{5^2}\times \frac{1}{1-\frac{1}{5}}$ [Sum of infinite GP $=\frac{a}{1-r}$]

$\Rightarrow \frac{11}{10}$

$\therefore 8k=11$