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

Answer: (C)

Let $S_{n}=\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}$
$=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\ldots \ldots \ldots+\frac{1}{3 n-1}-\frac{1}{3 n+2}\right]$
$=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right]$
$=\frac{n}{6 n+4}$