Express (2n+1)(2n+3)(2n+5) (4n-3)(4n-1) in factorials

[VinnyW]

Homework Statement

Express (2n+1)(2n+3)(2n+5)...(4n-3)(4n-1) in terms of factorials

Homework Equations

n!=n(n-1)!

The Attempt at a Solution

I know (2n+1)+(2n+3)+⋯+(4n−1)=∑2n−1+2k, where k starts as k = 1 and increases to infinity.
Then I was stuck. I am trying to learn maths on my own but it is getting more difficult.

 

[ehild]

Homework Statement

Express (2n+1)(2n+3)(2n+5)...(4n-3)(4n-1) in terms of factorials

Homework Equations

n!=n(n-1)!

The Attempt at a Solution

I know (2n+1)+(2n+3)+⋯+(4n−1)=∑2n−1+2k, where k starts as k = 1 and increases to infinity.
Then I was stuck. I am trying to learn maths on my own but it is getting more difficult.

How would you write B = (2n+1) (2n+2)(2n+3) (2n+4)(2n+5)...(4n-3) (4n-2)(4n-1) (4n) in terms of factorials? How is B related to the original expression A=(2n+1)(2n+3)(2n+5)...(4n-3)(4n-1)?

 

[HallsofIvy]

It might help you to know that a product of even numbers, 2(4)(6)...(2n-2)(2n), is equal to [2(1)][2(3)]...[2(n-1)][2(n)]= 2^n n!.

And a product of odd numbers, 1(3)(5)...(2n+1), is equal to \frac{1(2)(3)(4)...(2n-1)(2n)(2n+1)}{2(4)(6)...(2n)}= \frac{(2n+1)!}{2^n n!}

 

[epenguin]

Good news is this is/was pretty easy, bad news is you are/were stuck on it. You can hardly afford to be stuck on easy questions if you are on self-study. I recommend you get hold of Polya's "How to solve it" which is short and costs next to nothing.

It boils down to about five recommendations for when you are stuck.

The relevant one here is: take a simple example!

Write down your original question formula for some n, say 3 or 4. Write down B for the same n.
Write down n! for the same choice of n. Write down (4n - 1)! for that n. Write down (2n - 1)! Write down n!, write 2n! You should soon see what you have to do. And that that your Σ is far more advanced than needed for this question.

Tell us the answer. (If not I will not try to help you another time.)