Question

The sum of $n$ terms of the series$\frac{1}{1\cdot 3\cdot 5}+\frac{1}{3\cdot 5\cdot 7}+\frac{1}{5\cdot 7\cdot 9}+....$is

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

Answer 1

Answer: (B)

If $a_1,a_2,a_3...a_n\in A.P$., then $S_n$ of the series
$=\frac{1}{a_1{a}_2...a_{r-1}{a}_r}+\frac{1}{a_2{a}_3....a_{r+1}}+...+\frac{1}{a_n{a}_{n+1}...a_{n+r-1}}$
$=\frac{1}{\left(r-1\right)\left(a_2-a_1\right)}\left[\frac{1}{a_1{a}_2...a_{r-1}}-\frac{1}{a_{n+1}...a_{n+r-1}}\right]$
Here $r=3,a_2-a_1=3-1=2$
$\therefore$ Required Sum $=\frac{1}{2\cdot 2}\left[\frac{1}{1\cdot 3}-\frac{1}{\left(2n+1\right)\left(2n+3\right)}\right]$
$=\frac{1}{4}\left[\frac{4n^2+8n}{3\left(2n+1\right)\left(2n+3\right)}\right]=\frac{n\left(n+2\right)}{3\left(2n+1\right)\left(2n+3\right)}$
Alternative Solution : (Using difference method)
Here $t_n=\frac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}$
Let $V_n=\frac{1}{\left(2n+1\right)\left(2n+3\right)}$
$\therefore V_{n-1}=\frac{1}{\left(2n-1\right)\left(2n+1\right)}$
Now $V_n-V_{n-1}=\frac{1}{\left(2n+1\right)}\left[\frac{1}{2n+3}-\frac{1}{2n-1}\right]$
$=\frac{-4}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}$
$V_n-V_{n-1}=-4t_n$
$\therefore t_n=\frac{1}{4}\left[V_{n-1}-V_n\right]$
Putting $n=1,2,\mathrm{3......},$ then adding
$t_1+t_2+t_3+....t_n=S_n=\frac{1}{4}\left[V_0-V_n\right]$
$=\frac{1}{4}\left[\frac{1}{3}-\frac{1}{\left(2n+1\right)\left(2n+3\right)}\right]=\frac{n\left(n+2\right)}{3\left(2n+1\right)\left(2n+3\right)}$
Note: $V_0$ is obtained from
$V_{n-1}=\frac{1}{\left(2n-1\right)\left(2n+1\right)}$ by putting $n=1$
$\therefore V_0=\frac{1}{3}$