Binomial Theorem
Coefficient Conditions — Finding a+b+c
nta_pyq_2026_jan
Grade 11

Question:

If the coefficient of $x$ in the expansion of $(ax^2+bx+c)(1-2x)^{26}$ is $-56$ and the coefficients of $x^2$ and $x^3$ are both zero, then $a+b+c$ is equal to:
1500
1300
1403
1483

Step-by-Step Solution

Key Concept: In $(1-2x)^{26}$: $T_0=1$, $T_1=-52x$, $T_2=1300x^2$, $T_3=-20800x^3$. Set up equations from coeff of $x=-56$, coeff of $x^2=0$, coeff of $x^3=0$.
$c=3,b=100,a=1300$. $a+b+c=1403$.
Correct Answer: 3

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