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 equals to -

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

Answer 1

Answer: (A)

$\frac{13}{36}$

$T_n=\frac{1}{n^2+(n-2)}n=3,4,5,\mathrm{6....}\infty \Rightarrow T_n=\frac{1}{n^2+2n-n-2}=\frac{1}{(n+2)(n-1)}\Rightarrow T_n=\frac{1}{3}\left[\frac{(n+2)-(n-1)}{(n+2)(n-1)}\right]\Rightarrow T_n=\frac{1}{3}[\frac{1}{n-1}-\frac{1}{n+2}]$

So,

$S_{\infty }={\sum }_{n=3}^{\infty }{T}_n=\frac{1}{3}{\sum }_{n=3}^{\infty }[\frac{1}{n-1}-\frac{1}{n+2}]=\frac{1}{3}[\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\frac{1}{5}-\frac{1}{8}+\frac{1}{6}+.....\infty ]$

So, we see that all the terms will get cancelled except $\frac{1}{2},\frac{1}{3},\frac{1}{4}$

$\therefore S_{\infty }=\frac{1}{3}(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})S_{\infty }=\frac{1}{3}\left(\frac{6+4+3}{12}\right)S_{\infty }=\frac{13}{36}$