If the coefficient of $x$ in the expansion of $\left(a x^{2}+b x+c\right)(1-2 x)^{26}$ is -56 and the coefficients of $\mathrm{x}^{2}$ and $\mathrm{x}^{3}$ are both zero, then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Binomial Theorem

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

Answer: (C)

With

(1 - 2 x)^{26} = \sum_{k} (k 26) (- 2 x)^{k}

, the coefficient of

x

(a x^{2} + b x + c) (1 - 2 x)^{26}

b - 52 c = - 56

. The coefficient of

x^{2}

a - 52 b + 1300 c = 0

and that of

x^{3}

- 52 a + 1300 b - 20800 c = 0

. Solving these gives

c = 3, b = 100, a = 1300

, so

a + b + c = 1403

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option