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. $\frac{7}{9}$
• B. $\frac{7}{8}$
• C. $\frac{9}{7}$
• D. $\frac{9}{8}$

Answer 1

Answer: (C)

The general term in the expansion of $(3+ax{)}^9$ is
$T_{r+1}={}^9{C}_r{3}^{9-r}{a}^r{x}^r$
For coefficient of $x^2$, put $r=2$.
$\therefore T_{2+1}={}^9{C}_2{3}^{9-2}{a}^2{x}^2$
$\Rightarrow$ Coefficient of $x^2={}^9{C}_2{3}^7{a}^2....$(i)
For coefficient of $x^3$, put $r=3$.
$\therefore T_{3+1}={}^9{C}_3{3}^{9-3}{a}^3{x}^3$
$={}^9{C}_3{3}^6{a}^3{x}^3$
$\Rightarrow$ Coefficient of $x^3={}^9{C}_3{3}^6{a}^3....$(ii)
Given, coefficient of $x^2=$ Coefficient of $x^3$
${}^9{C}_2{3}^7{a}^2={}^9{C}_3{3}^6{a}^3\text{ [from Eqs. (i) and (ii)] }$
$\Rightarrow \frac{9\times 8}{2}\times 3\times 1=\frac{9\times 8\times 7}{6}\times 1\times a$
$\Rightarrow \frac{3}{2}=\frac{7}{6}\times a\Rightarrow a=\frac{9}{7}$