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

$$\begin{gathered} =(1+\mathrm{ax}+{\mathrm{bx}}^2){(1-2\mathrm{x})}^{18} \\ =(1+\mathrm{ax}+{\mathrm{bx}}^2)(1-{18}_{{\mathrm{c}}_1}(2\mathrm{x})+{18}_{{\mathrm{c}}_2}{\left(2\mathrm{x}\right)}^2\dots \dots \end{gathered}$$

cofficient of ${\mathrm{x}}^3$

$$\begin{gathered} ={18}_{{\mathrm{c}}_1}\left(2\mathrm{b}\right)+\mathrm{a}\times {18}_{{\mathrm{c}}_2}\left(4\right)-{18}_{{\mathrm{c}}_3}\left(8\right)=0 \\ 34\mathrm{a}=\mathrm{b}+\frac{34\times 16}{3} —\left(1\right) \end{gathered}$$

cofficient of ${\mathrm{x}}^4$

$$\begin{gathered} {18}_{{\mathrm{c}}_4}\left(4\right)-{18}_{{\mathrm{c}}_3}\left(2\mathrm{a}\right)+{18}_{{\mathrm{c}}_2}\left(\mathrm{b}\right)=0 — \left(2\right) \\ \mathrm{from} \left(1\right) \& \left(2\right) \\ \mathrm{a}=16, \mathrm{b}=\frac{272}{3} \end{gathered}$$