Question

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

• A. $(16,\frac{251}{3})$
• B. $(14,\frac{251}{3})$
• C. $(14,\frac{272}{3})$
• D. $(16,\frac{272}{3})$

Answer 1

The correct option is D $(16,\frac{272}{3})$

Given expression is $(1+ax+b{x}^2){(1-2x)}^{18}$

$=(1+ax+b{x}^2)\left(1{-}^{18}{C}_1\left(2x\right){+}^{18}{C_2{\left(2x\right)}^2-}^{18}{C}_3{\left(2x\right)}^3+...{.}^{18}{C}_{18}{\left(2x\right)}^{18}\right)$

Now, coefficient of $x^3$

$={-}^{18}{C}_1\left(2b\right)+a{\times }^{18}{C_2\left(4\right)-}^{18}{C}_3\left(8\right)=0$

$\therefore 34a=b+\frac{34\times 16}{3}...\left(i\right)$

Similarly, a coefficient of $x^4$

${}^{18}{C}_4\left(4\right){-}^{18}{C}_3\left(2a\right){+}^{18}{C}_2\left(b\right)=0...\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, $a=16$ and $b=\frac{272}{3}$