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JEE Main202621 January 2026Morning ShiftMathematicsBinomial TheoremActual

If the coefficient of x in the expansion of (a x²+b x+c )(1-2 x)²⁶ is -56 and the coefficients of x² and x³ are both zero, then a + b + c is equal to :

Options

  1. A1500
  2. B1300
  3. C1403
  4. D1483

Correct answer

C. 1403

Step-by-step solution

In (1-2x)²⁶ : T₀ = 1 , T₁ = -52 , T₂ = 1300 , T₃ = -20800 . Coefficient of x : b - 52c = -56 ...(1) Coefficient of x^2 : a - 52b + 1300c = 0 ...(2) Coefficient of x^3 : -52a + 1300b - 20800c = 0 ...(3) From (1): b = 52c - 56 . Substituting in (2): a = 1404c - 2912 . Substituting in (3): -26208c + 78624 = 0 c = 3 . Thus b = 100 , a = 1300 . a + b + c = 1300 + 100 + 3 = 1403 .

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