1. **Problem statement:** We need to find the values of $a$ and $b$ such that the coefficients of $x^3$ and $x^4$ in the expansion of $\left(1 + ax + bx^2\right)(1 - 2x)^{18}$ are both zero. 2. **Step 1:** Expand $(1-2x)^{18}$ using the binomial theorem: $$ (1-2x)^{18} = \sum_{k=0}^{18} {18 \choose k} (1)^{18-k} (-2x)^k = \sum_{k=0}^{18} {18 \choose k} (-2)^k x^k $$ 3. **Step 2:** We want to find the coefficients of $x^3$ and $x^4$ in the product: $$ (1 + ax + bx^2) \cdot \sum_{k=0}^{18} {18 \choose k} (-2)^k x^k $$ 4. **Step 3:** The coefficient of $x^n$ in the product is the sum of coefficients from multiplying terms that add powers to $n$: $$ \text{Coefficient of } x^n = \sum_{i=0}^2 \text{coefficient of } x^i \text{ in } (1+ax+bx^2) \times \text{coefficient of } x^{n-i} \text{ in } (1 - 2x)^{18} $$ 5. **Step 4:** Write the coefficient expressions explicitly: - Coefficient of $x^3$: $$ C_3 = 1 \times {18 \choose 3}(-2)^3 + a \times {18 \choose 2}(-2)^2 + b \times {18 \choose 1}(-2)^1 $$ - Coefficient of $x^4$: $$ C_4 = 1 \times {18 \choose 4}(-2)^4 + a \times {18 \choose 3}(-2)^3 + b \times {18 \choose 2}(-2)^2 $$ 6. **Step 5:** Calculate the binomial coefficients: - ${18 \choose 1} = 18$ - ${18 \choose 2} = \frac{18\times17}{2} = 153$ - ${18 \choose 3} = \frac{18\times17\times16}{6} = 816$ - ${18 \choose 4} = \frac{18\times17\times16\times15}{24} = 3060$ 7. **Step 6:** Substitute these and powers of $-2$: - $(-2)^1 = -2$ - $(-2)^2 = 4$ - $(-2)^3 = -8$ - $(-2)^4 = 16$ 8. **Step 7:** Write out $C_3$ and $C_4$: $$ C_3 = 1 \times 816 \times (-8) + a \times 153 \times 4 + b \times 18 \times (-2) = -6528 + 612a - 36b $$ $$ C_4 = 1 \times 3060 \times 16 + a \times 816 \times (-8) + b \times 153 \times 4 = 48960 - 6528a + 612b $$ 9. **Step 8:** Set $C_3 = 0$ and $C_4 = 0$: $$ -6528 + 612a - 36b = 0 \quad (1) $$ $$ 48960 - 6528a + 612b = 0 \quad (2) $$ 10. **Step 9:** Solve equation (1) for $b$: $$ 612a - 36b = 6528 \Rightarrow -36b = 6528 - 612a \Rightarrow b = \frac{612a - 6528}{36} = 17a - 181.333... $$ 11. **Step 10:** Substitute $b = 17a - \frac{544}{3}$ into equation (2): $$ 48960 - 6528a + 612(17a - \frac{544}{3}) = 0 $$ $$ 48960 - 6528a + 10404a - 612 \times \frac{544}{3} = 0 $$ 12. **Step 11:** Simplify the constants: $$ 10404a - 6528a = 3876a $$ $$ 612 \times \frac{544}{3} = 612 \times 181.333... = 110976 $$ 13. **Step 12:** Plug into equation: $$ 48960 + 3876a - 110976 = 0 $$ $$ 3876a - 62016 = 0 $$ $$ 3876a = 62016 $$ $$ a = \frac{62016}{3876} = 16 $$ 14. **Step 13:** Find $b$: $$ b = 17 \times 16 - \frac{544}{3} = 272 - \frac{544}{3} = \frac{816 - 544}{3} = \frac{272}{3} $$ 15. **Final answer:** The pair $(a, b)$ is $\left(16, \frac{272}{3}\right)$ which corresponds to option B.