Q44 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
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 $x^{2}$ and $x^{3}$ are both zero, then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to :
- A1500
- B1300
- Correct answerC1403
- D1483
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): $-26208c + 78624 = 0 \Rightarrow c = 3$.
Thus $b = 100$, $a = 1300$.
$a + b + c = 1300 + 100 + 3 = 1403$.
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): $-26208c + 78624 = 0 \Rightarrow c = 3$.
Thus $b = 100$, $a = 1300$.
$a + b + c = 1300 + 100 + 3 = 1403$.
Hence, the correct answer is option: 'C'.