Inverse Trigonometry
Telescoping cot-inverse sum
MJAT_TS6_P2
Grade 12
Question:
$\displaystyle\sum_{n=1}^\infty\cot^{-1}\!\left(\frac{(2n^2+2n+1)(n^2+n+1)}{n^4+2n^3+2n^2+2n+2}\right)=\sec^{-1}\!\left(\frac{5}{\lambda}\right)$. Then $\lambda$ equals:
A) 1
B) 2
C) $\dfrac{1}{2}$
D) $\sqrt{2}$
Step-by-Step Solution
Key Concept: The sum telescopes: $\cot^{-1}((n+1)^2)-\cot^{-1}(n^2)$ type. Sum $=\lim_{n\to\infty}\cot^{-1}((n+1)^2+...)-\cot^{-1}(1+1)=\pi/2-\cot^{-1}2=\tan^{-1}(1/2)$.
$\lambda=\mathbf{2}$. Answer: **B**.
Correct Answer: B