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Question #40525

Difficulty: 8/10 | 🟢 Free

The Question

Let In=∫2∞dx(x−1)2(x+1)nI_n = \int_2^\infty \frac{dx}{(x-1)^2 (x+1)^n} for n∈Nn \in \mathbb{N}. If Sn=4In−4In−1+In−2S_n = 4I_n - 4I_{n-1} + I_{n-2} for n≥3n \ge 3, evaluate the infinite series ∑n=3∞Sn\sum_{n=3}^\infty S_n.

Options

  • A.

    ln⁡(32)−13\ln\left(\frac{3}{2}\right) - \frac{1}{3}

  • B.

    ln⁡(32)−12\ln\left(\frac{3}{2}\right) - \frac{1}{2}

  • C.

    ln⁡3−1\ln 3 - 1

  • D.

    ln⁡(43)−13\ln\left(\frac{4}{3}\right) - \frac{1}{3}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Even Multiplicity Roots|Reduction Formulae|Partial Fractions Decomposition

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to navigate rational integrals featuring even multiplicity roots coupled with high powers.

The primary trap lies in attempting Partial Fraction Decomposition or recursive Integration by Parts.

Because of the even root (x−1)2(x-1)^2, standard IBP creates boundary term sign errors and recursive loops that do not easily close, leading to a 'Lethal Collision' when evaluating the infinite series.

The elegant bypass is recognizing that a Möbius substitution (x=1+t1−tx = \frac{1+t}{1-t}) maps the rational function to a binomial form, perfectly neutralizing the even multiplicity root, collapsing the recurrence algebraically, and revealing a standard logarithmic Taylor series.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLinear Inequalities - Class 11
ConceptEven Multiplicity Roots
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree