Question

The value of the sum $\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3}+\frac{1}{6^2+4}$.....$\infty$ is equal to

• A. $\frac{13}{36}$
• B. $\frac{12}{36}$
• C. $\frac{15}{36}$
• D. $\frac{18}{36}$

Answer 1

The correct option is A $\frac{13}{36}$

$\text{Solve:}\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3}+\frac{1}{6^2+4}+\dots \infty$

$\text{Its General term is}$

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

$a_n=\frac{1}{3}\left(\frac{3}{(n+1)(n+4)}\right)$

$\Rightarrow a_n=\frac{1}{3}[\frac{1}{n+1}-\frac{1}{n+4}]$

$\Rightarrow S_n={\sum }_{n=1}^{\infty }{a}_n$

Sn is the required Sum of Given series

$\Rightarrow S_n=\frac{1}{3}[(\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{5}-\frac{1}{8})+\dots$

$\Rightarrow S_n=\frac{1}{3}[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}]$ all other terms

are cancel each other

$\Rightarrow S_n=\frac{1}{3}\times \frac{6+4+3}{12}=\frac{13}{36}$

$\Rightarrow S_n=\frac{13}{36}$