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Question #40691

Difficulty: 8/10 | 🟢 Free

Let $S = \sum_{n=1}^{99} (-1)^n \tan^{-1}\left( \frac{2n+1}{1 - n - n^2} \right)$ . Find the value of $S$ .

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Method of Differences|Variable Transformation Substitution|The Sum-Difference Formula Constraints

This question tests the domain constraints of the inverse trigonometric sum-difference formulas.

Students often blindly apply $\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}(x) - \tan^{-1}(y)$ without checking if $xy > -1$ .

Here, setting $x=n$ and $y=-(n+1)$ results in $xy < -1$ , which requires a $-\pi$ correction factor to stay within the principal value branch.

Ignoring this shifts the final sum by $\pi$ , leading directly to the trap option.