Question

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

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

Answer 1

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

This can be written as
$(1-2x{)}^{18}+ax(1-2x{)}^{18}+b{x}^2(1-2x{)}^{18}$
Hence coefficient of $x^4$
$={}^{18}{C}_4{2}^4-a{}^{18}{C}_3{2}^3+b{}^{18}{C}_2{2}^2$
$=0$
Hence ${}^{18}{C}_4{2}^2-a{}^{18}{C}_3{2}^1+b{}^{18}{C}_2=0$ ...(i)
The coefficient of $x^3$
$=-{}^{18}{C}_3{2}^3+a{}^{18}{C}_2{2}^2-b{}^{18}{C}_1{2}^1=0$
$-{}^{18}{C}_3{2}^2+a{}^{18}{C}_2{2}^1-b{}^{18}{C}_1=0$ ..(ii)
Solving the above equations, we get
$a=16$ and $b=\frac{272}{3}$