Question

The sum of series ${\tan }^{-1}\left(\frac{4}{1+3\times 4}\right)+{\tan }^{-1}\left(\frac{6}{1+8\times 9}\right)+{\tan }^{-1}\left(\frac{8}{1+15\times 16}\right)+\dots \dots \infty$ is equal to

• A. ${\cot }^{-1}2$
• B. ${\tan }^{-1}2$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$T_n={\tan }^{-1}\left(\frac{2n+2}{1+\left(n^2+2n\right)\cdot (n+1{)}^2}\right)={\tan }^{-1}\left(\frac{(n+1)(n+2)-n(n+1)}{1+(n+1)(n+2)\cdot n(n+1)}\right)$
$T_n=\left({\tan }^{-1}((n+1)(n+2))-{\tan }^{-1}(n(n+1))\right)$
$S_n=\left({\tan }^{-1}((n+1)(n+2))-{\tan }^{-1}2\right)$
$\therefore \sum _{n=1}^{\infty }{T}_n=\left(\frac{\pi }{2}-{\tan }^{-1}2\right)={\cot }^{-1}2$.