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General Binomial Series

Mediummathematics

Treating the binomial theorem for a negative index, the coefficient of x^3 in the series expansion of (1 - x)^{-2} valid for |x| < 1 is determined to be which value?

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About This Question

Subject
mathematics
Chapter
binomial theorem and its simple applications
Topic
general binomial series
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drillnegative-indexbinomial-seriesconvergencecoefficient-extraction

Solution

Correct Answer:

For a negative integer index the binomial theorem generalises to an infinite series whose coefficients follow the rising-factorial pattern, a simple application that JEE Advanced includes under binomial series. The expansion (1 - x)^{-2} = sum_{r=0}^{\infty} (r + 1) x^r, since (1 - x)^{-n} = sum_{r} C(n + r - 1, r) x^r and for n = 2 this gives C(r + 1, r) = r + 1. Therefore the coefficient of x^3 corresponds to r = 3, giving r + 1 = 4. Hence the required coefficient is 4. Option 3 mistakenly uses r instead of r + 1. Option 6 applies the formula for (1 - x)^{-3} whose coefficients are C(r + 2, 2). Option 1 treats the series as a simple geometric one with all unit coefficients, valid only for (1 - x)^{-1}. The convergence condition |x| < 1 ensures the infinite series is meaningful. Plausibility check: differentiating the geometric series 1/(1 - x) = sum x^r gives 1/(1 - x)^2 = sum (r + 1) x^r, and the x^3 coefficient is indeed 4, confirming the negative-index expansion.

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

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