Before You Begin

Question #40662

Difficulty: 8/10 | 🟢 Free

The Question

Let $a_r = \sum_{k=0}^r \binom{2024}{k} \binom{2025}{r-k}$ for $0 \le r \le 4049$ . Define the polynomial $P(x) = \sum_{r=0}^{4049} a_r x^r$ and the function $f(x) = \ln(P(x))$ for $|x| < 1$ . If $g(x) = \int_0^x f(t) dt$ and $d_k$ denotes the coefficient of $x^k$ in the Maclaurin series expansion of $g(x)$ , find the value of $\left\lfloor \sum_{k=2}^{4050} |d_k| \right\rfloor$ , where $\lfloor \cdot \rfloor$ denotes the greatest integer function.

Options

A.
4047
B.
4048
C.
4049
D.
4050

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

Vandermonde's Identity|Logarithmic Series|Logarithmic Differentiation|Integral Part and Fractional Part

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize Vandermonde's Identity hidden within the coefficients of a polynomial before applying logarithmic expansion and integration.

The primary trap lies in attempting to directly differentiate or integrate the composite logarithmic function $\ln(P(x))$ without first reducing $P(x)$ to $(1+x)^{4049}$ , which leads to an insurmountable algebraic mess.

Furthermore, the 'Integral Part' constraint acts as a rigorous filter; students who incorrectly approximate the telescoping sum or fail to extract the exact fractional part $\frac{1}{4050}$ will easily fall for the off-by-one trap options (4049 or 4047).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Vandermonde's Identity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free