Question

If $S$ is the sum of the first 10 terms of the series ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)+{\tan }^{-1}\left(\frac{1}{21}\right)+\dots$, then $\tan (S)$ is equal to:

• A. $\frac{5}{11}$
• B. $-\frac{6}{5}$
• C. $\frac{10}{11}$
• D. $\frac{5}{6}$

Answer 1

Answer: (D)

$S={\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)+\dots$
$S={\tan }^{-1}\left(\frac{2-1}{1+1.2}\right)+{\tan }^{-1}\left(\frac{3-2}{1+2\times 3}\right)+{\tan }^{-1}$
$\left(\frac{4-3}{1+3\times 4}\right)+\dots +{\tan }^{-1}\left(\frac{11-10}{1+10\times 11}\right)$
$S=\left({\tan }^{-1}2-{\tan }^{-1}1\right)+\left({\tan }^{-1}3-{\tan }^{-1}2\right)+$
$\left({\tan }^{-1}4-{\tan }^{-1}3\right)+\dots \dots +\left({\tan }^{-1}(11)-{\tan }^{-1}(10)\right)$
$S={\tan }^{-1}11-{\tan }^{-1}1={\tan }^{-1}\left(\frac{11-1}{1+11}\right)$
$\tan (S)=\frac{11-1}{1+11\times 1}=\frac{10}{12}=\frac{5}{6}$