If coefficient of x3 and x4 in the expansion of 1+ax+bx2(1−2x)18 in powers of x are both zeros, then (a,b) is equal to

# If coefficient of ${x}^{3}$ and ${x}^{4}$ in the expansion of $\left(1+ax+b{x}^{2}\right)\left(1-2x{\right)}^{18}$ in powers of x are both zeros, then $\left(a,b\right)$ is equal to

1. A

$\left(14,\frac{251}{3}\right)$

2. B

$\left(14,\frac{272}{3}\right)$

3. C

$\left(16,\frac{272}{3}\right)$

4. D

$\left(16,\frac{251}{3}\right)$

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### Solution:

$S=\left(1+ax+b{x}^{2}\right)\left(1-2x{\right)}^{18}$

Coefficient of ${x}^{3}$ in the expansion of S is

Divide by  to obtain

$\frac{544}{3}-17a+b=0$               (1)

Similarly, coefficient of is

Divide by  to obtain

$80-\frac{32}{3}a+b=0$                 (2)

Subtract (2) from (1) to obtain

$\frac{304}{3}-\frac{19}{3}a=0⇒a=16$

From $b=17×16-\frac{544}{3}=\frac{272}{3}$

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