Question

The sum of infinte terms of the series $\frac{1}{1\cdot 4\cdot 7}+\frac{1}{4\cdot 7\cdot 10}+\cdots +\frac{1}{(3n-2)(3n+1)(3n+4)}$ is

• A. $\frac{1}{28}$
• B. $\frac{1}{24}$
• C. $\frac{1}{14}$
• D. $\frac{1}{12}$

Answer 1

Answer: (B)

$\frac{1}{24}$

$S_n=\frac{1}{1\cdot 4\cdot 7}+\frac{1}{4\cdot 7\cdot 10}+\cdots +\frac{1}{(3n-2)(3n+1)(3n+4)}$
Here $t_n$ = $\frac{1}{(3n-2)(3n+1)(3n+4)}$
Let $V_n=\frac{1}{(3n+1)(3n+4)}$ and $V_{n-1}=\frac{1}{(3n-2)(3n+1)}$
Substituting $n=0$, we get $V_0=\frac{1}{4}$
$V_n-V_{n-1}=6t_n$
$\Rightarrow t_n=\frac{1}{6}[V_{n-1}-V_n]$
Substituting $n=1,2,3$, ... then adding $t_1+t_2+t_3+...+t_n=S_n=\frac{1}{6}[V_0-V_n]$
$=\frac{1}{6}[\frac{1}{4}-\frac{1}{(3n+1)(3n+4)}]$
Therefore, $S_{\infty }=\underset{n\to \infty }{lim}{S}_n=\underset{n\to \infty }{lim}\frac{1}{6}[\frac{1}{4}-\frac{1}{(3n+1)(3n+4)}]=\frac{1}{24}$
Ans: B