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:

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

Answer 1

Answer: (D)

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