Question

The value of $\frac{1}{(1+a)(2+a)}+\frac{1}{(2+a)(3+a)}+\frac{1}{(3+a)(4+a)}+....+\infty$ is, (where, a is a constant).

• A. $\frac{1}{1+a}$
• B. $\frac{2}{1+a}$
• C. $\infty$
• D. None of these

Answer 1

The correct option is A $\frac{1}{1+a}$

$\frac{1}{(1+a)(2+a)}+\frac{1}{(2+a)(3+a)}+\frac{1}{(3+a)(4+a)}+....+\infty$
nth term of series
$T_n=\frac{1}{(n+a)(n+1+a)}=\frac{1}{n+a}-\frac{1}{n+1+a)}$
$T_1=\frac{1}{1+a}-\frac{1}{2+a}$
$T_2=\frac{1}{2+a}-\frac{1}{3+a}$
$T_3=\frac{1}{3+a}-\frac{1}{4+a}$
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$T_{n-1}=\frac{1}{n-1+a}-\frac{1}{n+a}$
$T_n=\frac{1}{n+a}-\frac{1}{n+1+a}$
$\therefore S_n=T_1+T_2+T_3+.....+T_n$
$=\frac{1}{1+a}-\frac{1}{n+1+a}$
$=\frac{n}{(1+a)(n+1+a)}$
$S_n=\frac{1}{(1+a)(1+\frac{1}{n}+\frac{a}{n})}$
$S_{\infty }=S_n,$ when $n\to \infty$
$\therefore S_{\infty }=\frac{1}{1+a}$.