Question
If the coefficient of x in the expansion of (a x^ 2 +b x+c )(1-2 x)^ 26 is -56 and the coefficients of x^ 2 and x^ 3 are both zero, then a + b + c is equal to :
JEE Main
Mathematics
Binomial Theorem
2026
JEE Main 2026 (21 January Shift 1)
JEE Main Mathematics Question (2026) — Solution
Options
- A. 1500
- B. 1300
- C. 1403
- D. 1483
Answer
C. 1403
Step-by-step solution
In (1-2x)^ 26 : T_0 = 1, T_1 = -52, T_2 = 1300, T_3 = -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.
Related: Mathematics — Binomial Theorem · All PYQ Banks