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Question #41420

Difficulty: 8/10 | 🟢 Free

The Question

Let Ar=∑k=0r(nk)(n−kr−k)A_r = \sum_{k=0}^r \binom{n}{k} \binom{n-k}{r-k} for 0≤r≤n0 \le r \le n. For an integer n≥2n \ge 2, define the function fn(x)f_n(x) for x>−12x > -\frac{1}{2} by fn(x)=∫0x∑r=2nArtrt2(2t+1)dtf_n(x) = \int_0^x \frac{\sum_{r=2}^n A_r t^r}{t^2(2t+1)} dt Let gn(x)=fn(x)−2(n−1)ln⁡(2x+1)g_n(x) = f_n(x) - 2(n-1)\ln(2x+1). Let SS be the set of all integers n≥2n \ge 2 such that gn(x)g_n(x) is a constant function on (−12,∞)\left(-\frac{1}{2}, \infty\right). Find the value of ∑n∈Sn+g4(1)\sum_{n \in S} n + g_4(1).

Options

  • A.

    14

  • B.

    18

  • C.

    20

  • D.

    24

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

Partial Fractions Decomposition|Properties of Binomial Coefficients|Coefficient of x^r in Expansion|Constant Function Logic

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to extract a polynomial from a complex binomial sum and perform advanced partial fraction decomposition.

The primary trap is the 'Constant Function Logic' distractor.

Students often assume that subtracting the logarithmic term 2(n−1)ln⁡(2x+1)2(n-1)\ln(2x+1) from the integral leaves a constant zero function, mistakenly thinking the numerator fully cancels out.

However, they miss the 'binomial residue'—the remaining polynomial (1+2x)n−1−1−2(n−1)xx2\frac{(1+2x)^{n-1} - 1 - 2(n-1)x}{x^2} which only vanishes for n=2n=2.

Recognizing this residue is crucial to correctly identifying the set SS and evaluating g4(1)g_4(1).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptPartial Fractions Decomposition
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree