Use the Principle of Mathematical Induction to prove that (2𝑛)! /(2n)𝑛! is odd for all positive integers.

For n=1, we have

$\frac{(2 n) !}{2 ! n !}=\frac{2 !}{2 ! !}=1 \text { is odd }$

Hence the result is true for n=1 (base case)

Let it be true for n=K.

Now for n=k+1

$\begin{aligned} \frac{(2 k+2) !}{2^{k+1} \cdot(k+1) !} &=\frac{(2 k+2)(2 k+1)(2 x) !}{2 \cdot 2^{k} \cdot(k+1) \cdot k !} \\ &=\frac{(2 k) !}{2^{k} \cdot k !} \cdot \frac{2(k+1) \cdot(2 k+1)}{2(k+1)} \\ &=\frac{(2 k) !}{2^{k} \cdot k !} \cdot(2 k+1) \\ &\text{where}\ \frac{(2 k) !}{2^{k} \cdot k !} ,(2 k+1)\ \text{are odd} \end{aligned}$

Hence the result is true for n=k+1.

Thus the result is true for all +ve integers by mathematical induction.