Question

The sum of the series $\frac{1}{3\times 7}+\frac{1}{7\times 11}+\frac{1}{11\times 15}+........to\infty$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{9}$
• D. $\frac{1}{12}$

Answer 1

Answer: (D)

$T_n=\frac{1}{\left[3+\left(n-1\right)4\right]\left[7+\left(n-1\right)4\right]}$

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

$=\frac{1}{4}\left[\frac{1}{4n-1}-\frac{1}{4n+3}\right]$

$\therefore T_1=\frac{1}{4}\left[\frac{1}{3}-\frac{1}{7}\right],T_2=\frac{1}{4}\left[\frac{1}{7}-\frac{1}{11}\right]$

$T_3=\frac{1}{4}\left[\frac{1}{11}-\frac{1}{15}\right]$ and so on.

$\therefore S_{\infty }=\frac{1}{4}\left[\frac{1}{3}\right]=\frac{1}{12}$

[All other terms will cancel]