Question

Sum to n terms of the series $\frac{1}{(1+x)(1+2x)}+\frac{1}{(1+2x)(1+3x)}+\frac{1}{(1+3x)(1+4x)}+...$ is

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

Answer 1

Answer: (B)

$\frac{n}{(1+x)[1+(n+1\left)x\right]}$

If $t_r$ denotes the nth term of the series, then
$x{t}_r=\frac{x}{(1+rx)(1+(r+1\left)x\right)}=\frac{1}{1+rx}-\frac{1}{1+(r+1)x}$
$\Rightarrow x{\sum }_{r=1}^n{t}_r={\sum }_{r=1}^n[\frac{1}{1+rx}-\frac{1}{1+(r+1)x}]=\frac{1}{1+x}-\frac{1}{1+(n+1)x}=\frac{nx}{(1+x)(1+(n+1\left)x\right)}$
$\Rightarrow {\sum }_{r=1}^n{t}_r=\frac{n}{(1+x)[1+(n+1\left)x\right]}$