Question

Find whether the statement is true/false.

$1+\frac{1}{5}+\frac{3}{5^2}+\frac{5}{5^3}+.....\infty$ = $\frac{11}{8}$

• A. True
• B. False

Answer 1

The correct option is A True

Let

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

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

From $\left(1\right)-\left(2\right)$, we get

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

$\frac{4S}{5}=1+\frac{2}{5^2}+\frac{2}{5^3}+\frac{2}{5^4}+.....\infty$

$\frac{4S}{5}-1=\frac{2}{5^2}+\frac{2}{5^3}+\frac{2}{5^4}+.....\infty$

$\frac{4S}{5}-1=2(\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+.....\infty )$

$\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+.....\infty$ is an infinite G.P with

$a=\frac{1}{5^2}$ and $r=\frac{1}{5}$

sum of infinite G.P $=\frac{a}{1-r}$

So,

$\frac{4S}{5}-1=2\left(\frac{\frac{1}{5^2}}{1-\frac{1}{5}}\right)$

$\frac{4S}{5}-1=2\left(\frac{1}{20}\right)$

$\frac{4S}{5}-1=\left(\frac{1}{10}\right)$

$\frac{4S}{5}=\frac{11}{10}$

$4S=\frac{11}{2}$

$S=\frac{11}{8}$

Hence, this is the answer.