Question

The value of $(\tan )^{-1}\left(\frac{9}{19}\right)+\tan ^{-1}\left(\frac{9}{49}\right)+(\tan )^{-1}\left(\frac{9}{97}\right)+(\tan )^{-1}\left(\frac{9}{163}\right)+\ldots \ldots$ 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=\displaystyle\lim_{m \rightarrow \infty} \sum_{n=1}^m(\tan )^{-1}\left(\frac{9}{9 n^2+3 n+7}\right) $
$=\displaystyle\lim _{m \rightarrow \infty} \sum_{n=1}^m(\tan )^{-1}\left(\frac{1}{1+n^2+\frac{n}{3}-\frac{2}{9}}\right)$
$=\displaystyle\lim _{m \rightarrow \infty} \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) $
$=\displaystyle\lim _{m \rightarrow \infty} \sum_{n=1}^m\left[(\tan )^{-1}\left(n+\frac{2}{3}\right)-(\tan )^{-1}\left(n-\frac{1}{3}\right)\right]$