Question

The sum to $n$ terms of the series $\frac{1}{(1+x)(1+3x)}+\frac{1}{(1+3x)(1+5x)}+\frac{1}{(1+5x)(1+7x)}+\dots \dots$, where $n>5,x\ge 2$ is

• A. $\frac{n}{(1+x)[1+(2n+1\left)x\right]}$
• B. $\frac{1}{(1+x)[1+(2n+1\left)x\right]}$
• C. $\frac{(1+x)}{n[1+(2n+1\left)x\right]}$
• D. $\frac{1+(2n+1)x}{n(1+x)}$

Answer 1

The correct option is A $\frac{n}{(1+x)[1+(2n+1\left)x\right]}$

$T_1=\frac{1}{(1+x)(1+3x)}$
So, $T_1=\frac{1}{2x}[\frac{1}{(1+x)}-\frac{1}{(1+3x)}]$
$T_2=\frac{1}{(1+3x)(1+5x)}$
So, $T_2=\frac{1}{2x}[\frac{1}{(1+3x)}-\frac{1}{(1+5x)}]$
Let $T_r$ be the general term of the series
$\therefore T_r=\frac{1}{2x}[\frac{1}{(1+(2r-1\left)x\right)}-\frac{1}{(1+(2r+1\left)x\right)}]$
So,
$S_n=\sum T_r=T_1+T_2+T_3+\dots \dots +T_n$
$=\frac{1}{2x}[\frac{1}{1+x}-\frac{1}{(1+(2n+1\left)x\right)}]=\frac{n}{(1+x)[1+(2n+1\left)x\right]}$