Question

The sum of series $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\dots$ up to $n$ terms is

• A. $\frac{n}{3(2n+3)}$
• B. $\frac{n}{(2n+3)}$
• C. $\frac{n}{(n+2)(n+4)}$
• D. none of these

Answer 1

The correct option is A $\frac{n}{3(2n+3)}$

For $n=1$, we have
$\frac{n}{3(2n+3)}=\frac{1}{3.5}=$ First term of the series
For $n=2$, we have
$\frac{n}{3(2n+3)}=\frac{2}{3.7}=\frac{1}{3.5}+\frac{1}{5.7}=$
Sum of the first two terms of the series.
Hence, option (A) is correct.
ALTER We have,
$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+$ to n terms
$=\frac{1}{2}[\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{(2n+1)(2n+3)}]$
$=[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+......+(\frac{1}{2n+1}-\frac{1}{2n+3})]$
$=\frac{1}{2}[\frac{1}{2}-\frac{1}{2n+3}]=\frac{n}{3(2n+3)}$