Question

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

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

Answer 1

The correct option is B $\frac{5}{6}$

$S={\tan }^{-1}\left(\frac{1}{1+1\times 2}\right)+{\tan }^{-1}\left(\frac{1}{1+2\times 3}\right)=\cdots$
$T_r={\tan }^{-1}\left(\frac{1}{1+r(r+1)}\right)$
$T_r={\tan }^{-1}(r+1)-{\tan }^{-1}r$
$T_1={\tan }^{-1}2-{\tan }^{-1}1$
$T_2={\tan }^{-1}3-{\tan }^{-1}2$
$T_3={\tan }^{-1}4-{\tan }^{-1}3$
$T_{10}={\tan }^{-1}11-{\tan }^{-1}10$
$\Rightarrow S=({\tan }^{-1}2-{\tan }^{-1}1)+({\tan }^{-1}3-{\tan }^{-1}2)+({\tan }^{-1}4-{\tan }^{-1}3)+\cdots +({\tan }^{-1}11-{\tan }^{-1}10)$
$\Rightarrow S={\tan }^{-1}11-{\tan }^{-1}1$
$\Rightarrow S={\tan }^{-1}\left(\frac{11-1}{1+11}\right)$
$\Rightarrow \tan S=\frac{10}{12}=\frac{5}{6}$