Question

$\frac {1} {2.5} +\frac {1} {5.8}+ \frac {1} {8.11}+.........\frac{1}{(3n-1)(3n+2)} $=

• A. $\frac {n}{6n+3}$
• B. $\frac {n}{6n-4}$
• C. $\frac {n+1}{6n+4}$
• D. $\frac {n}{6n+4}$

Answer 1

Answer: (D)

Given , $\frac{1}{2.5 } + \frac{1}{5.8} + \frac{1}{8.11} +... + \frac{1}{\left(3n-1\right)\left(3n+2\right)} $
$= \frac{1}{3} \left[\left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{5} -\frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{11}\right)+ .... + \left(\frac{1}{3n-1} - \frac{1}{3n+2}\right)\right] $
$ = \frac{1}{3} \left[\frac{1}{2} - \frac{1}{3n+2}\right]$
$ = \frac{3n+2-2}{6\left(3n+2\right)} = \frac{n}{6n+4} $