In the expansion of $(1+x)^{2 n}$, middle term is ${ }^{2 n} \mathrm{C}_n x^n$. Since, middle term is the greatest term.
$\therefore{ }^{2 n} \mathrm{C}_n x^n>{ }^{2 n} \mathrm{C}_{n-1} x^{n-1}$
$\text { and }{ }^{2 n} \mathrm{C}_r x^n>{ }^{2 n} \mathrm{C}_{n+1} x^{n+1}$
$\begin{aligned}
& \Rightarrow \quad x>\frac{{ }^{2 n} \mathrm{C}_{n-1}}{{ }^{2 n} \mathrm{C}_n} \text { and } x < \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}} \\
& \text { Hence, } x \in\left(\frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}, \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}\right)
\end{aligned}$
Hence, $x \in\left(\frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}, \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}\right)$
$\begin{aligned}
& x \in\left(\begin{array}{c}
\frac{(2 n) !}{(n-1) !(2 n-n+1)} \times \frac{n !(2 n-n) !}{(2 n) !} \\
\frac{(2 n) !}{n !(2 n-n) !} \times \frac{(n-1) !(2 n-n+1)}{(2 n) !}
\end{array}\right) \\
& x \in\left(\frac{n(n-1) ! n !}{n(n-1) !(n+1) n !}, \frac{(n+1) n !(n-1) !}{n(n-1) ! n !}\right) \\
& \therefore \quad x \in\left(\frac{n}{n+1}, \frac{n+1}{n}\right)
\end{aligned}$