Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Solution:

It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1}={ }^{n} C_{r} a^{n-t} b^{t}$.

Assuming that $x^{2}$ occurs in the $(r+1)^{\text {th }}$ term in the expansion of $(3+a x)^{9}$, we obtain

$\mathrm{T}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}(3)^{9-t}(\mathrm{ax})^{\mathrm{r}}={ }^{9} \mathrm{C}_{\mathrm{r}}(3)^{9-r} \mathrm{a}^{\mathrm{r}} \mathrm{x}^{\mathrm{r}}$

Comparing the indices of $x$ in $x^{2}$ and in $T_{r+1}$, we obtain

$r=2$

Thus, the coefficient of $x^{2}$ is

${ }^{9} \mathrm{C}_{2}(3)^{9-2} \mathrm{a}^{2}=\frac{9 !}{2 ! 7 !}(3)^{7} \mathrm{a}^{2}=36(3)^{7} \mathrm{a}^{2}$

Assuming that $x^{3}$ occurs in the $(k+1)^{\text {th }}$ term in the expansion of $(3+a x)^{9}$, we obtain

$\mathrm{T}_{\mathrm{k}+1}={ }^{9} \mathrm{C}_{\mathrm{k}}(3)^{9-\mathrm{k}}(\mathrm{ax})^{\mathrm{k}}={ }^{9} \mathrm{C}_{\mathrm{k}}(3)^{9-\mathrm{k}} \mathrm{a}^{\mathrm{k}} \mathrm{x}^{\mathrm{k}}$

Comparing the indices of $x$ in $x^{3}$ and in $T_{k+1}$, we obtain

$k=3$

Thus, the coefficient of $x^{3}$ is

${ }^{9} \mathrm{C}_{3}(3)^{9-3} \mathrm{a}^{3}=\frac{9 !}{3 ! 6 !}(3)^{6} \mathrm{a}^{3}=84(3)^{6} \mathrm{a}^{3}$

It is given that the coefficients of $x^{2}$ and $x^{3}$ are the same.

$84(3)^{6} a^{3}=36(3)^{7} a^{2}$

$\Rightarrow 84 a=36 \times 3$

$\Rightarrow \mathrm{a}=\frac{36 \times 3}{84}=\frac{104}{84}$

$\Rightarrow \mathrm{a}=\frac{9}{7}$

Thus, the required value of $a$ is $\frac{9}{7}$.