Question

Sum to $n$ term of the series $\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...$

• A. $\frac{n}{2(2n+1)}$
• B. $\frac{n}{4(2n+1)}$
• C. $\frac{n}{2n+1}$
• D. $\frac{n}{8(2n+1)}$

Answer 1

The correct option is B $\frac{n}{4(2n+1)}$

$S_n=\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+..$
$\Rightarrow 4S_n=\frac{6-2}{2.6}+\frac{10-6}{6.10}+\frac{14-10}{10.14}+...$
$=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+....-\frac{1}{2(2n+1)}$
$=\frac{1}{2}(1-\frac{1}{2n+1})=\frac{1}{2}\left(\frac{2n}{2n+1}\right)$
$\Rightarrow 4S_n=\frac{n}{2n+1}$ or $S_n=\frac{n}{4(2n+1)}$