If S is the sum of the first 10 terms of the series
Question:

If $\mathrm{S}$ is the sum of the first 10 terms of the series

$\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots$

then $\tan (S)$ is equal to :

1. $\frac{5}{11}$

2. $-\frac{6}{5}$

3. $\frac{10}{11}$

4. $\frac{5}{6}$

Correct Option: , 4

Solution:

$\mathrm{S}=\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\ldots$

$\mathrm{S}=\tan ^{-1}\left(\frac{2-1}{1+1.2}\right)+\tan ^{-1}\left(\frac{3-2}{1+2 \times 3}\right)+\tan ^{-1}$

$\left(\frac{4-3}{1+3 \times 4}\right)+\ldots .+\tan ^{-1}\left(\frac{11-10}{1+10 \times 11}\right)$

$S=\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+$

$\left(\tan ^{-1} 4-\tan ^{-1} 3\right)+\ldots . .+\left(\tan ^{-1}(11)-\tan ^{-1}(10)\right)$

$\mathrm{S}=\tan ^{-1} 11-\tan ^{-1} 1=\tan ^{-1}\left(\frac{11-1}{1+11}\right)$

$\tan (S)=\frac{11-1}{1+11 \times 1}=\frac{10}{12}=\frac{5}{6}$