Binomial Theorem

$\,{\left( {r + 2} \right)^{th}}$ term = ${}^{2n}{C_{r+1}}{\left( x \right)^r}$

And coefficient of $\,{\left( {r + 2} \right)^{th}}$ = ${}^{2n}{C_{r+1}}$

$3{r^{th}}$ term = ${}^{2n}{C_{3r - 1}}{\left( x \right)^{3r - 1}}$

And coefficient of $3{r^{th}}$ term = ${}^{2n}{C_{3r - 1}}$

According to the question,
${}^{2n}{C_{r+1}}$ = ${}^{2n}{C_{3r - 1}}$

$ \Rightarrow \left( {r + 1} \right) + \left( {3r - 1} \right) = 2n$

[As if ${}^n{C_p} = {}^n{C_q}$ then p + q = n]

$ \Rightarrow 4r = 2n$

$ \Rightarrow n = 2r$

Here in this expansion ${\left( {1 + x} \right)^{p + q}}$

The general term = ${T_{r + 1}} = {}^{p + q}{C_r}.{\left( x \right)^r}$

$\therefore$ ${x^p}$ will be present in the term = ${}^{p + q}{C_p}.{\left( x \right)^p}$

So coefficient of ${x^p}$ = ${}^{p + q}{C_p}$

And ${x^q}$ will be present in the term = ${}^{p + q}{C_q}.{\left( x \right)^q}$

$\therefore$ coefficient of ${x^q}$ = ${}^{p + q}{C_q}$

We know ${}^n{C_r}$ = ${}^n{C_{n - r}}$

$\therefore$ ${}^{p + q}{C_q}$ = ${}^{p + q}{C_{\left( {p + q} \right) - q}}$ = ${}^{p + q}{C_p}$

So coefficients of ${x^p}$ and ${x^q}$ are equal.