Question

The sum of $n$ terms of the series $\frac{1}{\mathrm{1.3.5}}+\frac{1}{\mathrm{3.5.7}}+\frac{1}{\mathrm{5.7.9}}+\cdots$ is

• A. $\frac{n(n+2)}{12(2n-1)(2n+3)}$
• B. $\frac{n(n+2)}{3(2n+1)(2n+3)}$
• C. $\frac{(n+1)(n+2)}{n(4n^2+1)}$
• D. $\frac{(n-1)(n+2)}{12(2n+1)(2n+3)}$

Answer 1

The correct option is B $\frac{n(n+2)}{3(2n+1)(2n+3)}$

$S_n=\frac{1}{\mathrm{1.3.5}}+\frac{1}{\mathrm{3.5.7}}+\frac{1}{\mathrm{5.7.9}}+\cdots$
Here $t_n$ = $\frac{1}{(2n-1)(2n+1)(2n+3)}$
Let $V_n=\frac{1}{(2n+1)(2n+3)}$ and $V_{n-1}=\frac{1}{(2n-1)(2n+1)}$
putting $n=0$, we get $V_0=\frac{1}{3}$
$V_n-V_{n-1}=4t_n$
$\Rightarrow t_n=\frac{1}{4}[V_{n-1}-V_n]$
Putting n=1, 2, 3, ... then adding $t_1+t_2+t_3+...+t_n=S_n=\frac{1}{4}[V_0-V_n]$
$=\frac{1}{4}[\frac{1}{3}-\frac{1}{(2n+1)(2n+3)}]=\frac{n(n+2)}{3(2n+1)(2n+3)}$
Ans: B