Inverse Trigonometry
Telescoping sum of cot⁻¹ series
MMTS_Full_Test_11
Grade 12
Question:
If $\displaystyle\sum_{n=1}^{\infty} \cot^{-1}\!\left(2 + \frac{n(n+1)}{2}\right) = \tan^{-1} a$, then $a$ is equal to
Step-by-Step Solution
Key Concept: Convert $\cot^{-1}(2+\frac{n(n+1)}{2})$ into a telescoping form using $\tan^{-1}\frac{1}{1+xy} = \tan^{-1}x - \tan^{-1}y$.
$S_\infty = \lim_{N\to\infty}\left(\tan^{-1}\frac{N+1}{2} - \tan^{-1}\frac{1}{2}\right) = \frac{\pi}{2} - \tan^{-1}\frac{1}{2} = \cot^{-1}\frac{1}{2} = \tan^{-1}2$. Hence $a=2$.
Correct Answer: (B) 2