If the coefficient of x^ 3 and x^ 4 in the expansion of (1 + ax +...

We have,&

$\left(1 + ax + bx^{2}\right) \left(1 - 2x\right)18$

\left(1 + ax + bx^{2}\right)\left(1 - C1 18 2x + C2 18 \left(2x^{2}\right) - C3 18 \left(2x\right)^{3} + C4 18 \left(2x\right)^{4} \ldots\right)$</p><p>Coefficient of$x^{3}$is$- C3 18 \left(2\right)^{3}$+ a \cdot C2 18 \left(2\right)^{2}$ - b^{18}{C}_{1}\left(2\right)$</p><p>and coefficient of$x^{4}$is$- C4 18 \left(2\right)^{4}$- C3 18 \left(2\right)$3a + C2 18 \left(2\right)$2b$

Coefficient of $x^{3} and x^{4}$ are zero.

$∴ - C3 18 \left(2\right)^{3} + C2 18$\left(2\right)^{2}\left(a\right)$- C1 18 \left(2\right)$ b = 0$</p><p>$⇒ $\dfrac{4 \times 17 \times 16}{3 \times 2} -$17a + b = 0... \left(i\right)$</p><p>and$80 - \dfrac{32}{3}$a + b = 0... \left(ii\right)$

Solving Eqs. $( i ) and ( ii )$, we get

$a = 16, b = \dfrac{272}{3}$