Question

$\frac{1}{\mathrm{1.4.7}}+\frac{1}{\mathrm{4.7.10}}+\frac{1}{\mathrm{7.10.13}}+\cdots \infty$ equals to

• A. $\frac{1}{12}$
• B. $\frac{1}{30}$
• C. $\frac{1}{24}$
• D. $\frac{1}{18}$

Answer 1

The correct option is C $\frac{1}{24}$

$S=\frac{1}{\mathrm{1.4.7}}+\frac{1}{\mathrm{4.7.10}}+\cdots$ n temrs
$T_r=\frac{1}{(3r-2)(3r+1)(3r+4)}$
$T_r=\frac{1}{6}[\frac{1}{(3r-2)(3r+1)}-\frac{1}{(3r+1)(3r+4)}]$
$T_1=\frac{1}{6}[\frac{1}{1.4}-\frac{1}{4.7}]$
$T_2=\frac{1}{6}[\frac{1}{4.7}-\frac{1}{7.10}]$
$T_n=\frac{1}{6}[\frac{1}{(3n-2)(3n+1)}-\frac{1}{(3n+1)(3n+4)}]$
$S_n=T_1+T_2+\cdots +T_n$
$S_n=\frac{1}{6}[\frac{1}{1.4}-\frac{1}{(3n+1)(3n+4)}]$
If $n\to \infty ,S_{\infty }=\frac{1}{24}$