Question

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

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

Answer 1

Answer: (A)

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