Question:

If $a=2$, then the sum of the infinite series $\cot^{-1}(2a^{-1}+a)+\cot^{-1}(2a^{-1}+3a)+\cot^{-1}(2a^{-1}+6a)+\cot^{-1}(2a^{-1}+10a)+\cdots$ is

$\dfrac{\pi}{4}$

$\dfrac{\pi}{2}$

$\dfrac{\pi}{3}$

$\dfrac{\pi}{6}$

Step-by-Step Solution

Key Concept: With $a=2$: terms are $\cot^{-1}(1+n(n+1))=\tan^{-1}\frac{1}{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}(n)$. Telescopes.

Sum $=\pi/4$.

Correct Answer: 1