Question

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.....$ upto $n$ terms $=$

• A. $\frac{n}{4n+6}$
• B. $\frac{1}{6n+4}$
• C. $\frac{n}{6n+4}$
• D. $\frac{n}{3n+7}$

Answer 1

The correct option is C $\frac{n}{6n+4}$

$2,5,8,....$

$t_n=2+(n-1)\ast 3=3n-1$

$5,8,....$

$t_n=5+(n-1)\ast 3=3n+2$

So series is $\frac{1}{2.5}+\frac{1}{5.8}+....+\frac{1}{(3n-1).(3n+2)}$

$\frac{1}{2.5}=\frac{1}{3}(\frac{1}{2}-\frac{1}{5})\frac{1}{5.8}=\frac{1}{3}(\frac{1}{5}-\frac{1}{8})...\frac{1}{(3n-1).(3n+2)}=\frac{1}{3}(\frac{1}{3n-1}-\frac{1}{3n+2})$

So series becomes

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

$=\frac{1}{3}(\frac{1}{2}-\frac{1}{5})+\frac{1}{3}(\frac{1}{5}-\frac{1}{8})+...\frac{1}{3}(\frac{1}{3n-1}-\frac{1}{3n+2})$

$=\frac{1}{3}(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....\frac{1}{3n-1}-\frac{1}{3n+2})=\frac{1}{3}(\frac{1}{2}-\frac{1}{3n+2})=\frac{3n}{6(3n+2)}=\frac{n}{6n+4}$