**Question:**

If the coefficients of $x^{2}$ and $x^{3}$ in the expansion of $(3+p x)^{9}$ are the same then prove that $\boldsymbol{P}=\frac{9}{7}$.

**Solution:**

To prove: that. If the coefficients of $x^{2}$ and $x^{3}$ in the expansion of $(3+p x)^{9}$ are

the same then $\boldsymbol{P}=\frac{9}{7}$.

Formula Used:

General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{\mathrm{n}} C_{r}=\frac{n !}{r !(n-r) !}$

Now, finding the general term of the expression, $(3+p x)^{9}$, we get

$T_{r+1}={ }^{9} C_{r} \times 3^{9-r} \times(p x)^{r}$

For finding the term which has ${ }^{x}{ }^{2}$ in it, is given by

$r=2$

Thus, the coefficients of $x^{2}$ are given by,

$T_{3}={ }^{9} C_{2} \times 3^{9-2} \times(\mathrm{px})^{2}$

$T_{3}={ }^{9} C_{2} \times 3^{7} \times p^{2} \times x^{2}$

For finding the term which has ${ }^{x}{ }^{2}$ in it, is given by

$r=3$

Thus, the coefficients of $x^{3}$ are given by,

$T_{3}={ }^{9} C_{3} \times 3^{9-3} \times(p x)^{3}$

$T_{3}={ }^{9} C_{3} \times 3^{6} \times p^{3} \times x^{3}$

As the coefficients of $x^{2}$ and $x^{3}$ in the expansion of $(3+p x)^{9}$ are the same.

${ }^{9} \mathrm{C}_{3} \times 3^{6} \times \mathrm{p}^{3}={ }^{9} \mathrm{C}_{2} \times 3^{7} \times \mathrm{p}^{2}$

${ }^{9} \mathrm{C}_{3} \times \mathrm{p}={ }^{9} \mathrm{C}_{2} \times 3$

$\frac{9 !}{3 ! \times 6 !} \times p=\frac{9 !}{2 ! \times 7 !} \times 3$

$\frac{9 !}{3 \times 2 ! \times 6 !} \times \mathrm{p}=\frac{9 !}{2 ! \times 7 \times 6 !} \times 3$

$\mathrm{p}=\frac{9}{7}$

Thus, the value of $p$ for which coefficients of $x^{2}$ and $x^{3}$ in the expansion of $(3+p x)^{9}$ are the same is $\frac{9}{7}$