Factorials

[angelina]

Use mathematical induction to prove that the product of the first n odd numbers, starting with 1, is given by (2n)!/2^n x n!
please note that the x n! is part of the fraction denominator, it is not multiplying the entire fraction

 

[tkhunny]

Well, do it.

1) Is it correct for some value of n?
2) Does one value of n imply the next value of n?
3) Parentheses would be far easier than your lengthy explanation of the denominator.

 

[galactus]

Use proper grouping symbols instead of explaining n!.

Just write (2n)!/(2^n*n!)

Anyway, but here is a start.

Note, the product of the n odd numbers is:

\(\displaystyle 1\cdot 3\cdot 5\cdot\cdot\cdot (2n-1)=\frac{(2n)!}{2^{n}n!}\)

Now, per the induction, the next term would be 2n+1.

Multiply both sides by 2n+1:

\(\displaystyle 1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)(2n+1)=\frac{(2n)!(2n+1)}{2^{n}n!}\)

Now, with some algebra, the right side should be equivalent to:

\(\displaystyle \frac{(2(n+1))!}{2^{n+1}(n+1)!}\)

Which shows the induction step.

Can you continue?.

 

[JeffM]

Use mathematical induction to prove that the product of the first n odd numbers, starting with 1, is given by (2n)!/2^n x n!
please note that the x n! is part of the fraction denominator, it is not multiplying the entire fraction

My response effectively is the same as that of galactus, but may perhaps be clearer.

What is the GENERAl method of mathematical induction. Let's say X is what you want to prove. 1st, ASSUME that X is true of some integer k. 2nd, PROVE on the basis of the prior assumption that X is true for (k + 1). 3rd, PROVE without using the first assumption that there is some integer m for which X is true. If you can do all three steps, you have proved that X is true for every integer n >= m.

So first you just assume that there is some k where the product of the first k odd numbers is [(2k)! / (k * 2^k]. (I like using a different letter for what you are assuming than the letter that goes into the proof because if you use the same letter it looks as though you are assuming what is to be proved. You are not.)

OK What is the kth odd number starting from one. It is (2k - 1), right?
So, what is the (k + 1)th odd number. It is [2(k + 1) - 1] = (2k + 1), right?

OK So now show that the product for (k + 1) odd numbers can be expressed in an identical form to the product of k odd numbers, with (k + 1) replacing k everywhere. Can you do that? If you can, you have done steps one and two. The idea is that if it is true for some integer k, it is also true for any integer bigger than k. Of course there may not be such a k. So the third step is required.

Now can you find integer m where you can show that the formula is true for m WITHOUT FIRST ASSUMING IT IS for k. Usually, you want to set m = 1, and prove whatever you are trying to prove for it because then it is true for all positive integers.

If you can, you have your proof.

Do you understand?