I need to know what 4!! equals and how you reached the answer

Thanks,
Heather

>I need to know what 4!! equals

4!! = 4·2

For any number n, n!! means the product of the decreasing sequence of natural numbers that starts with n and inlcudes every other number:

7!! = 7·5·3·1
8!! = 8·6·4·2
9!! = 9·7·5·3·1

> and how you reached the
> answer

What answer?

>I need to know what 4!! equals

n!! = (2^(n/2)) * (n/2)! for even values of n,

or n!! = n!/(n-1)!! for odd values of n

This means that 4!! = 8, using the first equation

> and how you reached the answer

example: 12!! = 12*10*8*6*4*2, or

(2)(6)*(2)(5)*(2)(4)*(2)(3)*(2)(2)*(2)(1), or

(2^6)*(6*5*4*3*2*1) = (2^6)(6!) = (2^(12/2))*(12/2)!

= (2^(n/2))*(n/2)!

Try this for other even numbers too, they all work.

For odd values of n, for example 11,

11!! = 11*9*7*5*3 = (11*10*9*8*7*6*5*4*3*2)/(10*8*6*4*2)

but the denominator is just (11-1)!! which can be found from the first equation. So the complete form for odd numbers is

n!/(n-1)!! and n-1 will be even if n is odd.

This isn't a formal proof but intuitively it works.

is there anywhere that I can find proof of this on the internet?

>is there anywhere that I can find proof of this on the
>internet?

A proof of what? It's a definition of n!!

n!!= n·(n-2)·(n-4)· ...

>is there anywhere that I can find proof of this on the
>internet?

I'm not sure, I just derived it. It's a fairly straight-forward concept so it shouldn't be too difficult to create a proof