1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How Do you reduce factorials

  1. Aug 19, 2008 #1
    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. jcsd
  3. Aug 19, 2008 #2
    I might be smoking something but:

    [tex]\frac{(2n+1)!}{(2n+3)!} = \frac{1*2*3*4*...*2n*(2n+1)}{1*2*3*4*...*2n*(2n+1)*(2n+2)*(2n+3)}[/tex]

    Doesn't it?
  4. Aug 19, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

    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.
  5. Aug 19, 2008 #4
    Thanks nicksauce for the formula, but I am still wondering how you would expand something like that (2n+1)!/(2n+3)! like Feldoh did.
  6. Aug 19, 2008 #5
    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.
  7. Aug 19, 2008 #6


    User Avatar

    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.
  8. Aug 19, 2008 #7
    Thank you all for helping
  9. Sep 5, 2010 #8
    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:

    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?
  10. Sep 6, 2010 #9
    hmm, i don't understand what you mean by expanding but

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

    when factor out (k+1)k! you get [tex][(k+1)k!(1+k+1)]-1[/tex] and you can complete the induction from here
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: How Do you reduce factorials
  1. How would you do this? (Replies: 8)