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JEE Main202628 January 2026Evening ShiftMathematicsBinomial TheoremActual

The sum of the coefficients of x⁴⁹⁹ and x⁵⁰⁰ in (1+x)¹⁰⁰⁰+x(1+x)⁹⁹⁹+x²(1+x)⁹⁹⁸+ +x¹⁰⁰⁰ is :

Options

  1. A1002 500
  2. B1002 501
  3. C1001 501
  4. D1000 501

Correct answer

A. 1002 500

Step-by-step solution

S = (1+x)¹⁰⁰⁰ + x(1+x)⁹⁹⁹ + x^2(1+x)⁹⁹⁸ + + x¹⁰⁰⁰ This is a geometric series with first term (1+x)¹⁰⁰⁰ , common ratio x 1+x , and 1001 terms. S = (1+x)¹⁰⁰⁰ 1 - ( x 1+x )¹⁰⁰¹ 1 - x 1+x = (1+x)¹⁰⁰¹ - x¹⁰⁰¹ Required sum = coefficient of x⁴⁹⁹ + coefficient of x⁵⁰⁰ in (1+x)¹⁰⁰¹ - x¹⁰⁰¹ = 1001 499 + 1001 500 = 1002 500

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