Yuri Y.

# Show that 2!*4!*6! *...* (2n)! >= ((n+1)!)^n

Mathematical Induction

2!*4!*6! *...* (2n)! >= ((n+1)!)n

I'm new here.
Thanks a lot for your helps and kindness

By: Tutor
5.0 (439)

Patient and Knowledgeable Tutor PoShan L.

Hi Yuri, I hope you find the answer helpful.  I have one suggestion, it will be helpful for us tutors and you, if you tell us what you have tried and where you get stuck.  I think you will have a higher chance of getting help and when you do, you will get an answer back quicker because we don't have to write a complete answer and can get straight to the point.  If you don't where where to start, it is okay, just say you need some help to get started.
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01/29/15

Yuri Y.

Thank you so much. I very appreciate this.
I find the answer very helpful but I'm still confused with some parts
(2(n+1))! =(2(n+1))*...*(n+3)*((n+1)+1)!

Why it's not (2(n+1))! = (2n+2) * (2n+1) * 2n * (2n-1) *.....

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01/29/15 PoShan L.

First of all, what you wrote is correct, HOWEVER, it is not useful, it doesn't go anywhere.  Now, why did I write it the way I wrote, the answer to that is, 1) it is correct, 2) it works.

Let me explain 1), I am assuming you agree that (2(n+1)) is the same thing as (2n + 2), so we are not disagreeing on the first part, now I want to point your attention to the ! in ((n+1)+1)!, in case you missed it, it is important that there is !.  If the ! is missing, it is wrong.  So what I did was kind of like writing

5! = 5*4*3!

do you agree that it is the same?  If you are not sure, look at this,

R.H.S = 5*4*3! = 5*4*3*2*1

Now if you agree, go back to look at

(2(n+1))! =(2(n+1))*...*(n+3)*((n+1)+1)!

do you see it this time?  If you still don't see it, go ahead and make up a number for n, and expand out all the terms in the !, then you will see that they are indeed equal.

Now, the motivation of doing it my way is because I see that the R.H.S still has a ! into it, so I know that I need to repackage the terms into a !.
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01/29/15

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