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

A
$(14, \frac{251}{3})$
B
$(14, \frac{272}{3})$
C
$(16, \frac{272}{3})$
D
$(16, \frac{251}{3})$

Solution:

$S = (1 + a x + b x^{2}) (1 - 2 x)^{18}$

$\begin{array}{l} = (1 + a x + b x^{2}) [1 +^{18} C_{1} (- 2 x) + \\ ^{18} C_{2} (- 2 x)^{2} +^{18} C_{3} (- 2 x)^{3} +^{18} C_{4} (- 2 x)^{4} + \dots] \end{array}$

Coefficient of $x^{3}$ in the expansion of S is $^{18} C_{3} (- 2)^{3} +$ $a (^{18} C_{2} (- 2)^{2}) +^{18} C_{1} (- 2) b = 0$

Divide by $^{18} C_{1} (- 2)$ to obtain

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

Similarly, coefficient of $x^{4}$ is

$^{18} C_{4} (- 2)^{4} + a (^{18} C_{3} (- 2)^{3}) +^{18} C_{2} (- 2)^{2} b = 0$

Divide by $^{18} C_{2} (- 2)^{2}$ to obtain

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

Subtract (2) from (1) to obtain

$\frac{304}{3} - \frac{19}{3} a = 0 \Rightarrow a = 16$

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

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

Solution:

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