Question

If the coefficient of $x^2$ and $x^3$ in the expansion of $(3+ax{)}^9$ are the same, then the value of a is

• A. $-\frac{7}{9}$
• B. $-\frac{9}{7}$
• C. $\frac{7}{9}$
• D. $\frac{9}{7}$

Answer 1

The correct option is D ($\frac{9}{7}$)

The $(r+1{)}^{\text{th}}$term of $(3+ax{)}^9$ is $T_{r+1}{=}^9{C}_r\left(3{)}^{9-r}\right(ax{)}^r$

$\Rightarrow T_{r+1}{=}^9{C}_r\left(3{)}^{9-r}\right(a{)}^r{x}^r$
Thus, the coefficient of $x^r$ is ${}^9{C}_r\left(3{)}^{9-r}\right(a{)}^r$
Substitute r=2 and r=3 to get the coefficient of $x^2$ and $x^3$ respectively.
Thus, the coefficient of $x^2$ is ${}^9{C}_2\left(3{)}^{9-2}\right(a{)}^2$
$=9C_2\left(3{)}^7\right(a{)}^2$
Similar way, the coefficient of $x^3$ is ${}^9{C}_3\left(3{)}^{9-3}\right(a{)}^3$
${=}^9{C}_3\left(3{)}^{9-3}\right(a{)}^3$
${=}^9{C}_3\left(3{)}^6\right(a{)}^3$
As the coefficients of $x^2$ and $x^3$ are equal.
$\Rightarrow 9C_2\left(3{)}^7\right(a{)}^2{=}^9{C}_3\left(3{)}^6\right(a{)}^3$
$\Rightarrow a=\frac{{}^9{C}_2}{{}^9{C}_3}\times 3$
$=\frac{\frac{9!}{7!2!}}{\frac{9!}{6!3!}}$
$=\frac{9!\times 3!\times 6!\times 3}{2!\times 7!\times 9!}$
$=\frac{3\times 2!\times 6!\times 3}{7\times 6!\times 2!}$
$\therefore a=\frac{9}{7}$