Question

The sum of infinite terms of the series $\frac{1}{1\cdot4\cdot7} + \frac{1}{4 \cdot7\cdot10} + ...+\frac{1}{\left(3n -2\right) \left(3n +1\right)\left(3n +4\right)}$ 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 $V_{n} = \frac{1}{\left(3n +1\right)\left(3n +4\right)}$
$V_{n -1} = \frac{1}{\left(3n -2\right)\left(3n +1\right)}$
$\therefore V_{0} = \frac{1}{4} \therefore S_{n} = \Sigma T_{n} = \frac{1}{6} \sum^{n}_{n=1} \left(V_{n -1}-V_{n}\right)$
$=\frac{1}{6} \left[ \frac{1}{4} -\frac{1}{\left(3n +1\right) \left(3n +4\right)}\right] = \frac{1}{24} \left[\frac{n \left(9n +15\right)}{\left(3n +1\right)\left(3n +4\right)}\right]$
$\therefore \lim_{n\to\infty}S_{n} = \frac{1}{24} \lim_{n\to\infty} \frac{n\left(9n +15\right)}{\left(3n +1\right)\left(3n +4\right)}=\frac{1}{24}$
Alternative Solution :
$S_{n} = \frac{1}{\left(a_{2} -a_{1}\right)} \frac{1}{\left(r -1\right)} \left[\frac{1}{a_{1}\cdot a_{2}}-\frac{1}{a_{n}a_{n +1}}\right]$
$= \frac{1}{3\cdot2}\left[\frac{1}{1\cdot4}-\frac{1}{\left(3n +1\right) \left(3n +4\right)}\right]$
$\therefore S_{n} = \frac{1}{6} \left[\frac{1}{4} -\frac{1}{\left(3n +1\right)\left(3n +4\right)}\right]$
$\therefore \lim_{n\to\infty} S_{n} = \frac{1}{6} \left[\frac{1}{4} -0\right] = \frac{1}{24}$