Question

The value of $(\tan {)}^{-1}\left(\frac{9}{19}\right)+{\tan }^{-1}\left(\frac{9}{49}\right)+(\tan {)}^{-1}\left(\frac{9}{97}\right)+(\tan {)}^{-1}\left(\frac{9}{163}\right)+\dots \dots$ is equal to

• A. $(\tan {)}^{-1}(3)$
• B. $(\tan {)}^{-1}\left(\frac{1}{3}\right)$
• C. $(\tan {)}^{-1}\left(\frac{2}{3}\right)$
• D. $(\tan {)}^{-1}\left(\frac{3}{2}\right)$

Answer 1

Answer: (D)

$S=\underset{m\to \infty }{\lim }\sum _{n=1}^m(\tan {)}^{-1}\left(\frac{9}{9n^2+3n+7}\right)$
$=\underset{m\to \infty }{\lim }\sum _{n=1}^m(\tan {)}^{-1}\left(\frac{1}{1+n^2+\frac{n}{3}-\frac{2}{9}}\right)$
$=\underset{m\to \infty }{\lim }\sum _{n=1}^m(\tan {)}^{-1}\left(\frac{\left(n+\frac{2}{3}\right)-\left(n-\frac{1}{3}\right)}{1+\left(n+\frac{2}{3}\right)\left(n-\frac{1}{3}\right)}\right)$
$=\underset{m\to \infty }{\lim }\sum _{n=1}^m\left[(\tan {)}^{-1}\left(n+\frac{2}{3}\right)-(\tan {)}^{-1}\left(n-\frac{1}{3}\right)\right]$