[JEE Main 2014] If the coefficients of $x^3$ and $x^4$ in $(1+ax+bx^2)(1-2x)^{18}$ are both zero, then $(a,b)$ is
If the coefficients of $x^3$ and $x^4$ in $(1+ax+bx^2)(1-2x)^{18}$ are both zero, then $(a,b)$ is
(a) $\left(14,\dfrac{272}3\right)$
(b) $\left(16,\dfrac{272}3\right)$
(c) $\left(16,\dfrac{251}3\right)$
(d) $\left(14,\dfrac{251}3\right)$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (b) $\left(16,\dfrac{272}3\right)$
Setting the two coefficients to $0$ gives $51a-3b=544$ and $-544a+51b=-4080$, solving to $a=16,\ b=\frac{272}3$.
JEE Main 2014 · Binomial Theorem — verified solution by the Vidaara Team.
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