If the coefficient of `x^(3)` and `x^(4)` in the expansion of `(1+ax+bx^(2)) (1-2x)^(18)` in power of x are both zero, then `(a,b)` is equal to

Options

A. `(16, (251)/(3))`
B. `(14, (251)/(3))`
C. `(14, (272)/(3))`
D. `(16, (272)/(3))`

Explanation

To find the coefficient of `x^(3)` and `x^(4)`, use the formula of coefficient of `x^(r)` in `(1-x)^(n)` is `(-1)^(r).^(n)C_(r)` and then simplify.
In expansion of `(1+ax+bx^(2))(1-2x)^(18)`.
Coefficient of `x^(3)="Coefficient of "x^(3)" in "(1-2x)^(18)`
`+"Coefficient of "x^(2)" in a "(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 `x^(3)=0`
`implies.^(18)C_(3).2^(3)+a.^(18)C_(2).2^(2)-b.^(18)C_(1).2=0`
`implies-(18xx17xx16)/(3xx2).8+a.(18xx17)/(2).2^(2)-b*18*2=0`
`implies17a-b=(34xx16)/(3)` . . . . (i)
Similarly coefficient of `x^(4)=0`
`implies.^(18)C_(4).2^(4)-a.^(18)C_(3)2^(3)+b.^(18)C_(2).2^(2)=0`
`therefore32a-3b=240` . . .(ii)
On solving Eqs. (i) and (ii), we get
`a-16,b-(272)/(3)`