Question

\(S=\tan^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan^{-1}\left(\frac{1}{n^2+3n+3}\right)+\ldots+\tan^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)\), then \(\tan S\) is equal to :

• A. \(\frac{20}{401+20n}\)
• B. \(\frac{n}{n^2+20n+1}\)
• C. \(\frac{20}{n^2+20n+1}\)
• D. \(\frac{n}{401+20n}\)

Hint:

Write each term as \(\tan^{-1}(k+1)-\tan^{-1}k\) so that the sum telescopes.

Answer 1

The Correct Option is C

Since \(n^2+n+1=1+n(n+1)\), \(\tan^{-1}\dfrac{1}{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}n\).
Similarly \(n^2+3n+3=1+(n+1)(n+2)\), giving \(\tan^{-1}(n+2)-\tan^{-1}(n+1)\), and so on, up to \(\tan^{-1}(n+20)-\tan^{-1}(n+19)\) for the last term \(1+(n+19)(n+20)\).
The sum telescopes: \(S=\tan^{-1}(n+20)-\tan^{-1}n\).
\(\tan S=\dfrac{(n+20)-n}{1+n(n+20)}=\dfrac{20}{n^2+20n+1}\).
Answer: (3) \(\frac{20}{n^2+20n+1}\)