# If S is the sum of the first 10 terms of the series

Question:

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

1. (1) $\frac{5}{6}$

2. (2) $\frac{5}{11}$

3. (3) $-\frac{6}{5}$

4. (4) $\frac{10}{11}$

Correct Option: 1

Solution:

$S=\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}+\ldots .$ upto 10 terms

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

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

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

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

$\operatorname{man}^{-1} 11-\tan ^{-1} 1=\tan ^{-1}\left(\frac{11-1}{1+11 \cdot 1}\right)=\tan ^{-1}\left(\frac{5}{6}\right)$

$\therefore \tan (S)=\frac{5}{6}$