Factorial Question

[lilshai]

Hello,
I know how to find factorials of numbers but how do you simplify factorials with variables?
i.e. how does (2n+1)! = (2n+1)(2n+2)?
Also, what would (2n+3)! equal?

Is there a general formula?

Thanks.

 

[stapel]

I'm sorry, but (2n + 1)! does not equal (2n + 1)(2n + 2). Instead, it equals (1)(2)(3)...(2n - 1)(2n)(2n + 1).

The ellipsis, the three-periods-in-a-row thing, is how one "does" factorials with variables, to indicate that "something goes in here, but I don't know specifically, because I don't know what the variable is".

Eliz.

 

[tkhunny]

(2n+1)! = (2n+1)(2n+2)

Let's see where this goes.

(2n+1)! = (2n+1)(2n+2)
(2n+1)*[(2n)]! = (2n+1)(2n+2)
(2n)! = (2n+2) <== That's enough.

n = 1 2! = 2+2?? 2 = 4 <== Whoops.

 

[lilshai]

I'm sorry, but (2n + 1)! does not equal (2n + 1)(2n + 2). Instead, it equals (1)(2)(3)...(2n - 1)(2n)(2n + 1).

The ellipsis, the three-periods-in-a-row thing, is how one "does" factorials with variables, to indicate that "something goes in here, but I don't know specifically, because I don't know what the variable is".

Eliz.

I'm sorry, you are right. I left out the other part of the equation. It was (2n)!/ (2n+2)! = 1/(2n+1)(2n+2). Basically, I am wondering how they got that.

How did they simplify the factorials?

 

[stapel]

They write out the expanded form, the way I showed you, and cancel off the duplicate portion. For instance:

. . .(n + 2)! / (n - 1)!

. . . . .= [(1)(2)(3)...(n - 2)(n - 1)(n)(n + 1)(n + 2)] / [(1)(2)(3)...(n - 2)(n - 1)]

. . . . .= (n)(n + 1)(n + 2)

...after cancelling the common "(1)(2)(3)...(n - 2)(n - 1)" part.

Eliz.

 

[lilshai]

They write out the expanded form, the way I showed you, and cancel off the duplicate portion. For instance:

. . .(n + 2)! / (n - 1)!

. . . . .= [(1)(2)(3)...(n - 2)(n - 1)(n)(n + 1)(n + 2)] / [(1)(2)(3)...(n - 2)(n - 1)]

. . . . .= (n)(n + 1)(n + 2)

...after cancelling the common "(1)(2)(3)...(n - 2)(n - 1)" part.

Eliz.

Hi, thanks for the response. I'm not sure of what method you are using...I am using the general formula for factorials which is n! = 1· 2· 3· · · (n − 2)(n − 1)n. I figured out my original question using this formula, however, when I try to use it using the example you did, I get (n+2)! = 1· 2· 3· · · (n + 2)(n + 1)n and for (n+1)! = 1· 2· 3· · · (n - 3)(n - 2)(n - 1). I just replace n with (n+2) or (n+1) in the general formula. Am I doing something wrong?

 

[stapel]

I'm not sure of what method you are using.

How is your expansion different from my expansion, that we're doing "different methods"?

Eliz.

 

[lilshai]

How is your expansion different from my expansion, that we're doing "different methods"?

Eliz.

oh ok, I see what you did. You included the previous terms in (n+2)! so that you could cancel the (n-3)(n-2)(n-1) terms. I guess it would be incorrect to write [n(n+1)(n+2)]/[(n-3)(n-2)(n-1)] which is what I had.