Question

If the coefficient of $x^3$ and $x^4$ in the expansion of $(1+ax+b{x}^2)(1-2x{)}^{18}$ in the 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,
$(1+ax+b{x}^2)(1-2x{)}^{18}$
$=(1-2x{)}^{18}+ax(1-2x{)}^{18}+b{x}^2(1-2x{)}^{18}$
$\text{Coefficient of}{x}^3={}^{18}{C}_3(-2{)}^3+a{}^{18}{C}_2(-2{)}^2+{}^{18}{C}_1(-2{)}^{1}b$
$=-\frac{544}{3}+17a-b=\mathrm{0...}\left(1\right)$
$\text{Coefficient of}{x}^4={}^{18}{C}_4(-2{)}^4+a{}^{18}{C}_3(-2{)}^3+{}^{18}{C}_2(-2{)}^{2}b$
$=80-\frac{32}{3}a+b=\mathrm{0...}\left(2\right)$
On adding equation $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow a=16,b=\frac{272}{3}$