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

Explanation

We have,
`(1 + ax + bx ^(2)) (1 - 2x )^(18)`
`(1 + ax = bx ^(2))sum_(r=0)^(18) ""^(18)C_(r) (-1)^(r) 2 ^(r) x^(r) `
`= sum _(r=0) ^(18) ""^(C_(r) (-1)^(r) 2^(r) x^(r) + sum _(r=0)^(18) ""^(18)C_(r) (-1)^(r) 2^(r) x^(r -1) + b sum_(r=0)^(18) ""^(18)C_(r) (-1)^(r) 2^(r) x^(r +2)`
The coefficient of `x^(3) and x^(4)` in the expansion of `( + ax + bx^(2))`
` (1 - 2x)^(18)` are both zero .
`therefore ""^(18)C_(3) (-1)^(3) 3^(3) + a ""^(18)C_(2) (-1)^(2) 2^(2) + b ""^(18)C_(1) (-1)^(1) 2^(1) = 0`
and, ` ""^(18)C_(4) (-1)^(4) 2^(4) + a ""^(18)C_(3) (-1)^(3) 2^(3) + b ""^(18)C_(2) (-1)^(2) 2^(2) = 0`
`rArr 51a - 3b = 144 and 32 a - 3b = 240`
` rArr a = 16, b = (272)/(3)` .