Question

The sum of infinite series $\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\frac{1}{63}\dots \dots$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A)

$\Theta S=\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\frac{1}{63}\dots .$
$=\frac{1}{3}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\dots \dots \right)$
$=\frac{2}{3}\left(\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\dots \dots \right)$
$=\frac{2}{3}\left(\sum _{n=2}^{\infty }\frac{1}{n(n+1)}\right)$
$=\frac{2}{3}\left(\sum _{n=2}^{\infty }\left(\frac{1}{n}-\frac{1}{n+1}\right)\right)=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}$