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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Binomial Theorem

If the coefficient of x in the expansion of (a x^ 2 +b x+c)(1-2 x)^...

If the coefficient of xx in the expansion of \left(a x^{2}+b x+c\right)(1-2 x)^{26}is−56andthecoefficientsofis -56 and the coefficients ofx^{2} and x^{3}arebothzero,thenare both zero, then\mathrm{a}+\mathrm{b}+\mathrm{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(1-2x)^{26}: T0=1T_0 = 1, T1=−52T_1 = -52, T2=1300T_2 = 1300, T3=−20800T_3 = -20800.
Coefficient of xx: b−52c=−56b - 52c = -56 ...(1)
Coefficient of x2x^2: a−52b+1300c=0a - 52b + 1300c = 0 ...(2)
Coefficient of x3x^3: −52a+1300b−20800c=0-52a + 1300b - 20800c = 0 ...(3)
From (1): b=52c−56b = 52c - 56. Substituting in (2): a=1404c−2912a = 1404c - 2912.
Substituting in (3): −26208c+78624=0⇒c=3-26208c + 78624 = 0 \Rightarrow c = 3.
Thus b=100b = 100, a=1300a = 1300.
a+b+c=1300+100+3=1403a + b + c = 1300 + 100 + 3 = 1403.
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