Binomial Theorem · Question 17

JEE Main 2026 (Online) 21st January Morning ShiftMCQMedium📈 Topic in 23% papers+4−1

If the coefficient of x in the expansion of (ax2 + bx + c)(1 - 2x)26 is -56 and the coefficients of x2 and x3 are both zero, then a + b + c is equal to :

1483
1300
1500
1403

Answer

(4) 1403

Solution

(1 - 2x)26 = r=026 26Cr(-2x)r = 1 - 52x + 1300x2 - 20800x3 +

26C1 2 = 52, 26C2 4 = 325 4 = 1300, 26C3 8 = 2600 8 = 20800

Multiplying by (ax2 + bx + c):

Coefficient of x: b - 52c = -56 ...(i)

Coefficient of x2: a - 52b + 1300c = 0 ...(ii)

Coefficient of x3: -52a + 1300b - 20800c = 0

a - 25b + 400c = 0 ...(iii)

(ii) - (iii): -27b + 900c = 0

b = 100c3

Put in (i): 100c3 - 52c = -56

- 56c3 = -56

c = 3, b = 100

From (ii): a = 52b - 1300c = 5200 - 3900 = 1300

a + b + c = 1300 + 100 + 3 = 1403

Answer: (4)

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