Question

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

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

Answer 1

Answer: (A)

$S=\frac{1}{\mathrm{1.3.5}}+\frac{1}{\mathrm{3.5.7}}+\frac{1}{\mathrm{5.7.9}}+\dots ..n$
$T_n=\frac{1}{(2n-1)(2n+1)(2n+3)}$
$\therefore T_n=\frac{1}{4}\left[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}\right]$
$T_1=\frac{1}{4}\left[\frac{1}{1.3}-\frac{1}{3.5}\right],T_2=\frac{1}{4}\left[\frac{1}{3.5}-\frac{1}{5.7}\right]$
$T_3=\left[\frac{1}{5.7}-\frac{1}{7.9}\right]$
$T_n=\left[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}\right]$
sum of all terms gives $S_n$
$S_n=\frac{1}{4}\left[\frac{1}{3}-\frac{1}{(2n+1)(2n+3)}\right]$