Question

If \(\cot^{-1}(\alpha) = \cot^{-1}2 + \cot^{-1}8 + \cot^{-1}18 + \cot^{-1}32 + \cdots\) upto 100 terms, then \(\alpha\) is :

• A. 1.02
• B. 1.03
• C. 1.01
• D. 1.00

Hint:

The \(n\)th term is \(\cot^{-1}(2n^2)\); write \(\tan^{-1}\frac{1}{2n^2} = \tan^{-1}\frac{1}{2n-1} - \tan^{-1}\frac{1}{2n+1}\) and telescope.

Answer 1

The Correct Option is C

Terms 2, 8, 18, 32, ... are \(2n^2\), so the \(n\)th term is \(\cot^{-1}(2n^2) = \tan^{-1}\dfrac{1}{2n^2}\).
Note \(\tan^{-1}\dfrac{1}{2n-1} - \tan^{-1}\dfrac{1}{2n+1} = \tan^{-1}\dfrac{\frac{2}{4n^2 - 1}}{1 + \frac{1}{4n^2 - 1}} = \tan^{-1}\dfrac{2}{4n^2} = \tan^{-1}\dfrac{1}{2n^2}\).
Sum over \(n = 1\) to 100 telescopes: \(\tan^{-1}1 - \tan^{-1}\frac{1}{201} = \tan^{-1}\dfrac{1 - \frac{1}{201}}{1 + \frac{1}{201}} = \tan^{-1}\dfrac{200}{202} = \tan^{-1}\dfrac{100}{101}\).
So \(\cot^{-1}\alpha = \tan^{-1}\frac{100}{101} \Rightarrow \alpha = \frac{101}{100} = 1.01\).
Answer: (3) 1.01