Question

If $ta{n}^{-1}\frac{1}{1+2}+ta{n}^{-1}\frac{1}{1+\left(2\right)\left(3\right)}+ta{n}^{-1}\frac{1}{1+\left(3\right)\left(4\right)}+....+ta{n}^{-1}\frac{1}{1+n(n+1)}=ta{n}^{-1}\theta$, then $\theta =$

• A. $\frac{n}{n+1}$
• B. $\frac{n+1}{n+2}$
• C. $\frac{n}{n+2}$
• D. $\frac{n-1}{n+2}$

Answer 1

The correct option is C $\frac{n}{n+2}$

Writing the general term, we get

${\tan }^{-1}\frac{1}{1+n(n+1)}$
$={\tan }^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right)$
$={\tan }^{-1}(n+1)-{\tan }^{-1}\left(n\right)$

Hence the series becomes
${\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)+...{\tan }^{-1}(n+1)-{\tan }^{-1}\left(n\right)$
$={\tan }^{-1}(n+1)-{\tan }^{-1}1$
$={\tan }^{-1}\left(\frac{n}{n+2}\right)$
Hence $\theta =\frac{n}{n+2}$