Question

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

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

Answer 1

Answer: (B)

$T_n=\frac{1}{6}\left[V_{n-1}-V_n\right]$ (By theory)
Where $V_n=\frac{1}{\left(3n+1\right)\left(3n+4\right)}$
$V_{n-1}=\frac{1}{\left(3n-2\right)\left(3n+1\right)}$
$\therefore V_0=\frac{1}{4}\therefore S_n=\Sigma T_n=\frac{1}{6}{\sum }_{n=1}^n\left(V_{n-1}-V_n\right)$
$=\frac{1}{6}\left[\frac{1}{4}-\frac{1}{\left(3n+1\right)\left(3n+4\right)}\right]=\frac{1}{24}\left[\frac{n\left(9n+15\right)}{\left(3n+1\right)\left(3n+4\right)}\right]$
$\therefore {\lim }_{n\to \infty }{S}_n=\frac{1}{24}{\lim }_{n\to \infty }\frac{n\left(9n+15\right)}{\left(3n+1\right)\left(3n+4\right)}=\frac{1}{24}$
Alternative Solution :
$S_n=\frac{1}{\left(a_2-a_1\right)}\frac{1}{\left(r-1\right)}\left[\frac{1}{a_1\cdot a_2}-\frac{1}{a_n{a}_{n+1}}\right]$
$=\frac{1}{3\cdot 2}\left[\frac{1}{1\cdot 4}-\frac{1}{\left(3n+1\right)\left(3n+4\right)}\right]$
$\therefore S_n=\frac{1}{6}\left[\frac{1}{4}-\frac{1}{\left(3n+1\right)\left(3n+4\right)}\right]$
$\therefore {\lim }_{n\to \infty }{S}_n=\frac{1}{6}\left[\frac{1}{4}-0\right]=\frac{1}{24}$