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

Answer 1

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

$(1+ax+b{x}^2)(1–2x{)}^{18}=1(1-2x{)}^{18}+ax(1-2x{)}^{18}+b{x}^2(1-2x{)}^{18}$
Coefficient of $x^3:(-2{)}^3{}^{18}{C}_3+a(-2{)}^2{}^{18}{C}_2+b(-2){}^{18}{C}_1=0$
$\frac{4\times (17\times 16)}{(3\times 2)}-2a\times \frac{17}{2}+b=0$----- (1)
Coefficient of $x^4:(-2{)}^4{}^{18}{C}_4+a(-2{)}^3{}^{18}{C}_3+b(-2{)}^2{}^{18}{C}_2=0$
$(4\times 20)-2a\times \frac{16}{3}+b=0$ ---- (2)
From equations (1) and (2), we get
$4(\frac{17\times 8}{3}-20)+2a(\frac{16}{3}-\frac{17}{2})=0$
$\Rightarrow$ a = 16
$\Rightarrow b=\frac{2\times 16\times 16}{3}-80=\frac{272}{3}$