Show that the middle term in the expansion of $(1+x)^{2n}$ is $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} 2^n x^n$,where $n$ is a positive integer.

Since $2n$ is even,the middle term of the expansion $(1+x)^{2n}$ is the $(\frac{2n}{2} + 1)^{\text{th}}$ term,i.e.,the $(n+1)^{\text{th}}$ term.
The $(n+1)^{\text{th}}$ term is given by:
$T_{n+1} = {}^{2n}C_n (1)^{2n-n} (x)^n = {}^{2n}C_n x^n = \frac{(2n)!}{n!n!} x^n$
$= \frac{2n(2n-1)(2n-2) \cdots 4 \cdot 3 \cdot 2 \cdot 1}{n!n!} x^n$
$= \frac{[1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n!n!} x^n$
$= \frac{[1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot 2^n [1 \cdot 2 \cdot 3 \cdots n]}{n!n!} x^n$
$= \frac{[1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot n!}{n!n!} 2^n x^n$
$= \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} 2^n x^n$