Question

If the coefficients of $x^2$ and $x^3$ in the expansion of $(3+ax{)}^9$are equal then the value of $a$ is:

• A. $\frac{3}{7}$
• B. $\frac{7}{3}$
• C. $\frac{7}{9}$
• D. $\frac{9}{7}$

Answer 1

The correct option is D

$\frac{9}{7}$

Explanation for the correct option.

Step 1: Information required for the solution

In an expansion of $(3+ax{)}^9$, the coefficients of $x^2$ and $x^3$ are equal.

We know that the general term in the expansion of $(a+b{)}^{\mathrm{n}}$ is given by:

${\mathbf{T}}_{\mathbf{r}\mathbf{+}\mathbf{1}}\mathbf{=}{{}^{\mathbf{n}}\mathbf{C}}_{\mathbf{r}}{\mathbf{a}}^{\mathbf{n}\mathbf{-}\mathbf{r}}{\mathbf{b}}^{\mathbf{r}}$

In the expansion of $(3+ax{)}^9$ the $\text{coefficient of}{x}^2=\text{coefficient of}{x}^3$ which can be written as

${{}^{9}C}_2{3}^{9-2}{\mathrm{a}}^2={{}^9\mathrm{C}}_3{3}^{9-3}{\mathrm{a}}^3$

Step 2: Calculation for the value of $a$

Use the above equation then,

$\begin{array}{rcl}a& =& \frac{{{}^{9}C}_2{3}^{9-2}}{{{}^{9}C}_3{3}^{9-3}}\\ & =& \frac{{{}^{9}C}_2{3}^7}{{{}^{9}C}_3{3}^6}\\ & =& \frac{{\displaystyle \frac{9·8}{2·1}}}{{\displaystyle \frac{9·8·7}{3·2·1}}}\left(3^{7-6}\right)\\ & =& \frac{3}{7}\left(3\right)\\ & =& \frac{9}{7}\end{array}$

Hence, option D is correct.