# How Do you reduce factorials

1. Aug 19, 2008

### viciousp

1. The problem statement, all variables and given/known data
This isn't a specific problem rather I don't know how to reduce factorials and this is giving me a hard time when I try the ratio test. For an example I'll use (2n+1)!/(2n+3)!

2. Relevant equations

3. The attempt at a solution
I attempt it by writing out some solutions in this case for n=0 to n=3 or so.
I start of writing it as (1*3*5*7)/(3*5*7*9) so it reduces down to 1/9 when n=3 so the incorrect reduced equation is 1/(2n+3). How should I approach problems like these?

2. Aug 19, 2008

### Feldoh

I might be smoking something but:

$$\frac{(2n+1)!}{(2n+3)!} = \frac{1*2*3*4*...*2n*(2n+1)}{1*2*3*4*...*2n*(2n+1)*(2n+2)*(2n+3)}$$

Doesn't it?

3. Aug 19, 2008

### nicksauce

Using the fact that (x)! = (x)(x-1)! (where x = 2n+3 in your case), you can easily show that that fraction reduces to the form that Feldoh's post suggests.

4. Aug 19, 2008

### viciousp

Thanks nicksauce for the formula, but I am still wondering how you would expand something like that (2n+1)!/(2n+3)! like Feldoh did.

5. Aug 19, 2008

### Alex6200

Well just remember that (2n + 1)! = (2n + 1) * 2n * (2n - 1) * (2n - 2)!

And then just cancel the common multiples out on the top and bottom of the fraction.

6. Aug 19, 2008

### Staff: Mentor

let x = (2n+3) and apply the formula nicksauce provided.

also if x! = x (x-1)!, then (x-1)! = (x-1) (x-2)! and so on . . . .

or alternatively, realize (x+1)! = (x+1) x!, then (x+2)! = (x+2) (x+1)! = (x+2)(x+1) x!.

Basically one expand the larger of the numerator or denominator until the common factorial appears.

7. Aug 19, 2008

### viciousp

Thank you all for helping

8. Sep 5, 2010

### greywolfmom

I have a question expanding off of this. I am working a mathematical induction problem. Original question is 1*1!+2*2!+...(k)*(k)!=(n+1)!-1, and n is greater than or equal to 1.
The first two steps I completed with little problems.

For my last step I need to prove that n=k+1
My problem so far:
1*1!+2*2!+...(k+1)*(k+1)!=((k+1)+1)!-1

LHS is now:
((k+1)!-1+(k+1)*(k+1)! My problem lies in reducing this to match my RHS ((k+1)+1)!-1
My guess is that expanding it would look like this (k+1)*k!-1+(k+1)*(k+1)*k!, how do I reduce from here, but more importantly why? I don't understand why it would be acceptable to factor out (k+1)! or (k+1)*k! from either the non expanded or expanded form.
If I factor out (k+1)! without expanding it first I get 1-1+(k+1)*1, and I get the same when I factor out (k+1)*k! from the expanded form. Any insight to where I am going wrong on reducing it?

9. Sep 6, 2010

### annoymage

hmm, i don't understand what you mean by expanding but

when factor out (k+1)! you should get $$[(k+1)!(1+k+1)]-1$$

when factor out (k+1)k! you get $$[(k+1)k!(1+k+1)]-1$$ and you can complete the induction from here