Aspire Faculty ID #18112 · Topic: JEE Main 2026 (28 January Evening Shift) · Just now
JEE Main 2026 (28 January Evening Shift)

The sum of coefficients of $ x^{499} $ and $ x^{500} $ in

Solution

$ S = (1 + x)^{1000} + x(1 + x)^{999} + x^2(1 + x)^{998} + \ldots + x^{1000} $

$ = (1 + x)^{1000} \left( 1 - \left( \frac{x}{1 + x} \right)^{1001} \right) $

$ = (1 + x)^{1000} - x^{1001} $

Required sum $ = {}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500} $

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