Answer:
Option D
Explanation:
To find the coefficient of x3 and x4, use the formula of coefficient of xf in (1-x)n is (-1)r nCr and then simplify.
In expansion of $(1+ax+bx^{2})$ (1-2x)18
Coefficient of x3 = Coefficient of x3 in (1-2x)18+ coefficient of x2 ina (1-2x)18 + Coefficient of x in b(1-2x)18.
= $-^{18}C_{3}.2^{3}+a^{18}C_{2}.2^{2}-b^{18}C_{1}.2$
Given, coefficient of x3 =0
=$^{18}C_{3}.2^{3}+a^{18}C_{2}.2^{2}-b^{18}C_{1}.2$ =0
$\Rightarrow$ $-\frac{18\times 17 \times 16}{3\times 2}.8+a.\frac{18\times17}{2}.2^{2}-b.18.2=0$
$\Rightarrow$ $ 17a-b=\frac{34\times16}{3}$ .......(i)
Simolarly coefficient of x4
$^{18}C_{4}.2^{4}-a^{18}C_{3}.2^{3}-b^{18}C_{2}.2^{2}=0$
$\therefore $ $32a-3b=240$ ..........(ii)
On solving Eqs (i) and (ii), we get,
a=16, $b=\frac{272}{3}$
