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Coefficient Extraction

Mediummathematics

A question requires the coefficient of x^5 in the product expansion of (1 + x)^4 multiplied by (1 + x)^6, where both factors are first combined sensibly.

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About This Question

Subject
mathematics
Chapter
binomial theorem and its simple applications
Topic
coefficient extraction
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drillcoefficient-extractionexponent-additionvandermondeproduct-of-binomials

Solution

Correct Answer:

Products of binomial powers with the same base are best merged by adding exponents before extracting any coefficient, a streamlined JEE Advanced technique that avoids tedious convolution. Since (1 + x)^4 (1 + x)^6 = (1 + x)^{4+6} = (1 + x)^{10}, the coefficient of x^5 is simply C(10, 5). Evaluating, C(10, 5) = 10! / (5! 5!) = 252. Hence the required coefficient is 252. Option 210 = C(10, 6) corresponds to the wrong power of x. Option 120 = C(10, 3) again targets a different term. Option 45 = C(10, 2) is far from the central term. The crucial simplification is recognising that multiplying the powers adds the exponents, collapsing the product into a single binomial. Alternatively, the Vandermonde identity sum over C(4, k)C(6, 5 - k) for k from 0 to 4 reproduces the same 252, cross-validating the merge and showing why the convolution of the two coefficient sequences agrees with the single merged coefficient. This equivalence is no accident: multiplying generating functions corresponds exactly to convolving their coefficient sequences, which is the algebraic content of the Vandermonde identity. Plausibility check: the coefficient of x^5 should be the central, hence largest, coefficient of (1 + x)^{10}, and 252 is indeed the maximum entry in that row, consistent with the symmetry of Pascal's triangle.

This medium difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of coefficient extraction. It appeared in the 2025 exam.

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