JEE Main 2026 Jan 21 Shift 1, Mathematics Q16: Introduction to Binomial Expansion

JEE Main2026Jan 21, Shift 1Mathematics

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

A
1300
B
1500
C
1403
D
1483

Solution

Let $T_{r} = (r 26) (- 2)^{r}$ denote the coefficient of $x^{r}$ in $(1 - 2 x)^{26}$ . Then collecting powers in the product:

Coeff. of $x$ : $b T_{0} + c T_{1} = b - 52 c = - 56$ .

Coeff. of $x^{2}$ : $a T_{0} + b T_{1} + c T_{2} = a - 52 b + 1300 c = 0$ .

Coeff. of $x^{3}$ : $a T_{1} + b T_{2} + c T_{3} = - 52 a + 1300 b - 20800 c = 0$ .

Solving the three linear equations: $a = 1300$ , $b = 100$ , $c = 3$ . So $a + b + c = 1403$ .

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JEE Main 2026 Jan 21 Shift 1, Mathematics Q16: Introduction to Binomial Expansion

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