Question

If the coefficients 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. (14, 272/3)
• B. (16, 272/3)
• C. (16, 251/3)
• D. (14, 251/3).

Answer 1

Answer: (B)

To find the coefficient of $x^3$ and $x^4$, use the formula of coefficient of $x^r$ in $(1-x{)}^n$ is $(-1{)}^r$ ${}^n{C}_r$and then simplify
. In expansion of $(1+ax+b{x}^2)(1-2x{)}^{18}$
Coefficient of $x^3$ = Coefficient of $x^3$ in $(1-2x{)}^{18}$
$+$ Coefficient of $x^2$ in a $(1-2x{)}^{18}$
$+$ Coefficient of $x$ in $b(1-2x{)}^{18}$
$={-}^{18}{C}_3.2^3+a{}^{18}{C}_a.2^2-b^{18}{C}_1.2$
Given, coefficient of $x^3=0$
$\Rightarrow {}^{18}{C}_3.2^3+a{}^{18}{C}_2.2^2-b{}^{18}{C}_1.2=0$
$\Rightarrow -\frac{18\times 17\times 16}{3\times 2}.8+a.\frac{18\times 17}{2}{2}^2-b.18.2=0$
$\Rightarrow 17a-b=\frac{34\times 16}{3}$....(i)
Similarly, coefficient of $x^4=0$
$\Rightarrow {}^{18}{C}_4.2^4-a.{}^{18}{C}_3{2}^3+b.{}^{18}{C}_2.2^2=0$
$\therefore 32a-3b=240$.......(ii)
On solving Eqs. (i) and (ii), we get
$a=16,b=\frac{272}{3}$