Question

Sum to n terms of the series ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)+...$ is

• A. ${\tan }^{-1}\left(\frac{n}{n+2}\right)$
• B. ${\tan }^{-1}\left(\frac{2n-1}{2n+2}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{3n}\right)$
• D. ${\tan }^{-1}\left(\frac{n}{n+1}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{n}{n+2}\right)$

The each term of the series can be written as $ta{n}^{-1}\frac{(n+1)-n}{1+n(n+1)}$
i.e, $=\sum \left(ta{n}^{-1}\right(n+1)-ta{n}^{-1}n)=ta{n}^{-1}(n+1)-ta{n}^{-1}1$
$=ta{n}^{-1}\frac{n}{n+2}$