factorial rules

Nikitin

hey. I've had no education in factorials specifically, but my professor is expecting us to already know this stuff...

In a problem where factorials are included, it is claimed (2n+2)! = (2n+2)*(2n+1)*(2n)!. Shouldn't it be (2n+2)!= (2n+2)*2n! ?

In addition, is there any difference between 2n! and (2n)! ?

arildno

1. "(2n+2)!= (2n+2)*2n! ?" No, why?

2. "In addition, is there any difference between 2n! and (2n)! ?"

2*n! is twice the value of n!
(2n!) is the factorial up to the number 2n

For example, 2*3! =12, whereas (2*3)!=6!=720

Mentallic

What is a factorial? n! is defined as being all the positive integers up to and including n being multiplied together, so n! = n(n-1)(n-2)...3*2*1

Ok, using this idea, what would (2n+2)! be? Well, first we multiply by (2n+2), then we reduce the value by 1 and multiply by that, so we multiply by ((2n+2)-1) = (2n+1).

Yes,
2n! = 2*(n!) = 2*(n(n-1)(n-2)...3*2*1)

while
(2n)! = (2n)(2n-1)(2n-2)...*3*2*1

Nikitin

1. Doesn't (2n+2)! = (2n+2)*(2(n-1)+2)*(2(n-2)+2)!

2. Ah yes. How could I miss that? damn, i guess i'm exhausted.

ah, OK. thanks 4 the help

Mentallic

I have a feeling you might be getting this mixed up with the techniques you learnt in Mathematical induction?

Let's give n a value, say, n=5

(2n+2)! = (2*5+2)! = 12! = 12*11*...*3*2

Now, 2(n-1)+2 = 2*4+2 = 10.
Notice how 10 is 2 less than 12, because we didn't take 1 away from the value 12, we took a value away from n, which is being multiplied by 2, so if we followed what you wrote we'd end up with 12! = 12*10*8*6*4*2

Essentially, if we have a linear equation an+b for some constants a and b, then [itex]an+b-1 \neq a(n-1)+b[/itex] unless a = 1.