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Question #40684

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the following infinite series: $S = \sum_{n=1}^{\infty} \tan^{-1}\left( \frac{(n-1)2^n}{(n^2 - 5n + 6)2^{2n+1} + 3(3n - 7)2^n + 10} \right)$

Options

A.
$-\frac{\pi}{4}$
B.
$\frac{3\pi}{4}$
C.
$\frac{\pi}{4}$
D.
$\frac{5\pi}{4}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Arithmetic-Geometric Progression (AGP)|Infinite Series involving Inverse Tangent|The Quadrant Adjustment Logic|Indeterminate Forms

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize and sum a telescoping series involving inverse trigonometric functions, while navigating a critical domain trap.

The core trap lies in the 'Quadrant Adjustment Logic' for the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}(\frac{x-y}{1+xy})$ .

Students habitually apply this formula without verifying the condition $xy > -1$ .

By embedding an Arithmetic-Geometric Progression (AGP) into the argument, the sequence of terms $x_n$ is forced to cross from negative to positive, specifically making $x_2 x_3 = -3 < -1$ .

This necessitates a $-\pi$ phase shift in the telescoping sum.

Failing to account for this shifts the final limit at infinity, leading directly to the distractor option $3\pi/4$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Arithmetic-Geometric Progression (AGP)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free