Question #284652

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


1
Expert's answer
2022-01-05T04:57:28-0500

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

For n=1n=1 we get that 2!21!=1\frac{2 !}{2 \cdot 1 !}=1 is odd.

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

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

It follows that (2(k+1))!2k+1(k+1)!=(2k)!(2k+2)(2k+1)2k2(k+1)k!=(2k)!2kk!(2k+1).\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 (2k)!2kk!\frac{(2k)!}{2^k k!} and 2k+12k+1 are odd, we conclude that (2k)!2kk!(2k+1)\frac{(2k)!}{2^k k!}(2k+1) is also odd. Therefore, we proved that (2(k+1))!2k+1(k+1)!\frac{(2(k+1))!}{2^{k+1} (k+1)!} is odd.

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


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