Answer to Question #284652 in Discrete Mathematics for triple

Question

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

Answer 1

Let us use the Principle of Mathematical Induction to prove that $\frac{(2n) !}{2^n n !}$ is odd for all positive integers.

For $n=1$ we get that $\frac{2 !}{2 \cdot 1 !}=1$ is odd.

Assume that the statement is true for $n=k,$ that is $\frac{(2k)!}{2^k k!}$ is odd.

Let us prove for $n=k+1.$

It follows that $\frac{(2(k+1))!}{2^{k+1} (k+1)!}=\frac{(2k)!(2k+2)(2k+1)}{2^k2(k+1) k!} =\frac{(2k)!}{2^k k!}(2k+1).$ Taking into account that $\frac{(2k)!}{2^k k!}$ and $2k+1$ are odd, we conclude that $\frac{(2k)!}{2^k k!}(2k+1)$ is also odd. Therefore, we proved that $\frac{(2(k+1))!}{2^{k+1} (k+1)!}$ is odd.

We conclude that by the Principle of Mathematical Induction  $\frac{(2n) !}{2^n n !}$ is odd for all positive integers.