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

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

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

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

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

answer is B.

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Detailed Solution

$\begin{array}{l} (1 + a x + b x^{2}) (1 - 2 x)^{18} \\ = (1 + a x + b x^{2}) [1 + {}^{18}C_{1} (- 2 x) + {}^{18}C_{2} (- 2 x)^{2} + {}^{18}C_{3} (- 2 x)^{3} + \dots] \\ Coefficient x^{3} is 0 \Rightarrow {}^{18}C_{3} (- 2)^{3} + a^{18} C_{2} (- 2)^{2} + b^{18} C_{1} (- 2) = 0 \end{array}$

$\begin{array}{l} \Rightarrow - 6528 + 612 a - 36 b = 0 \Rightarrow 51 a - 3 b = 544 \to (1) \\ Coefficient x^{4} is \Rightarrow {}^{18}C_{4} (- 2)^{4} + a^{18} C_{3} (- 2)^{3} + b^{18} C_{2} (- 2)^{2} = 0 \\ \Rightarrow 48960 - 6528 a + 612 b = 0 \\ \Rightarrow 32 a - 3 b = 240 \to (2) \\ (1) - (2) \Rightarrow 19 a = 304 \Rightarrow a = 16 \\ \Rightarrow 512 - 3 b = 240 \Rightarrow 3 b = 272 \Rightarrow b = \frac{272}{3} \end{array}$