Inverse Trigonometry
Telescoping series of cot⁻¹
Grade Class 12
Question:
The value of $\displaystyle\sum_{r=2}^{\infty} \cot^{-1}(r^2 - 5r + 7)$ is
$\dfrac{\pi}{4}$
$\dfrac{\pi}{2}$
$\dfrac{3\pi}{4}$
$\dfrac{5\pi}{4}$
Step-by-Step Solution
Key Concept: Rewrite $\cot^{-1}(r^2-5r+7)$ as $\tan^{-1}\frac{1}{(r-2)(r-3)+1} = \tan^{-1}(r-2)-\tan^{-1}(r-3)$ to telescope.
Telescoping: sum $=\frac{\pi}{2}-\tan^{-1}(-1)=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}$.
Correct Answer: 3