Solution to Use the Principle of Mathematical Induction to prove that ((2n) !)/(2^n n !)((2n)!)/(2 n n!) … - Sikademy
Author Image

Archangel Macsika

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

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

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.

Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-385-qpid-272