Sum to n terms of the series 1(1+x)(1+2x)+1(1+2x)(1+3x)+1(1+3x)(1+4x)+…is

If tr denotes the rth term of the series, thenxtr=x(1+rx)(1+(r+1)x)=11+rx−11+(r+1)x⇒ x∑r=1n tr=∑r=1n 11+rx−11+(r+1)x =11+x−11+(n+1)x =nx(1+x)(1+(n+1)x)⇒ ∑r=1n tr=n(1+x)[1+(n+1)x]

Sum to n terms of the series 1(1+x)(1+2x)+1(1+2x)(1+3x)+1(1+3x)(1+4x)+…is

If tr denotes the rth term of the series, thenxtr=x(1+rx)(1+(r+1)x)=11+rx−11+(r+1)x⇒ x∑r=1n tr=∑r=1n 11+rx−11+(r+1)x =11+x−11+(n+1)x =nx(1+x)(1+(n+1)x)⇒ ∑r=1n tr=n(1+x)[1+(n+1)x]

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

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

$\frac{x}{(1 + x) (1 + (n - 1) x)}$

none of these

$\frac{n x}{(1 + x) (1 + n x)}$

answer is B.

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Detailed Solution

If $t_{r}$ denotes the $r t h$ term of the series, then

$x t_{r} = \frac{x}{(1 + r x) (1 + (r + 1) x)} = \frac{1}{1 + r x} - \frac{1}{1 + (r + 1) x}$

$\begin{array}{l} \Rightarrow x \sum_{r = 1}^{n} t_{r} = \sum_{r = 1}^{n} [\frac{1}{1 + r x} - \frac{1}{1 + (r + 1) x}] \\ = \frac{1}{1 + x} - \frac{1}{1 + (n + 1) x} \\ = \frac{n x}{(1 + x) (1 + (n + 1) x)} \end{array}$
$\Rightarrow \sum_{r = 1}^{n} t_{r} = \frac{n}{(1 + x) [1 + (n + 1) x]}$