Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help on factorials

  1. Oct 10, 2006 #1
    can you please explain (step by step) how to factor the following:

    (n+1)! - 1 + (n+1)(n+1)!

    i have the answer, don't know how to get there!
     
  2. jcsd
  3. Oct 10, 2006 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    radou, that's absolutely wrong. You realized you just proved (n+1)!=1 for all n?
     
  4. Oct 10, 2006 #3

    radou

    User Avatar
    Homework Helper

    Yes, I just did. :biggrin:
     
  5. Oct 10, 2006 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's not fair! You can't delete your post like that! :P

    Getting back on topic:

    (n+1)! + (n+1)*(n+1)! - 1 = (n+1)!*(1 + n + 1) - 1.

    Can you go from there?
     
  6. Oct 10, 2006 #5
    thanks for the quick reply...
    You see that's where i get confused...how did you end up with (1 + n + 1)...
    Is (n+1)! = (n-1)(n)(n+1)....and so forth????!!
     
  7. Oct 10, 2006 #6

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    (n+1)! + (n+1)*(n+1)! = (n+1)!*1 + (n+1)!*(n+1). You factor (n+1)! out and are left with 1 + n + 1

    And yes, (n+1)! = (n+1)*n*(n-1)...
     
  8. Oct 10, 2006 #7
    does the factoring process of (n+1)! involve (n+1)! = (n+1)*n*(n-1)... ?
     
  9. Oct 10, 2006 #8
    [tex] (n+1)! = n!(n+1) [/tex]
     
  10. Oct 10, 2006 #9
    oh all right i see what you guys are coming at....would it have been easier if i had substituted any variable (say, h) for (n+1)!....? and then factored it...
    Btw, this is part of a mathematical induction....im trying to understand factorials better!!
    thank you guys!
     
  11. Oct 10, 2006 #10
    (this does not involve factorials anymore)....
    [k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4
    factor this out....
    What's the common factor? How did you get there? (ok i hope it doesnt require expanding the polynomials :p)
    Again, would it be easier if i substituted every (k+x) by a different variable, where (k+1) would equal to variable 'A', (k+2) = B, and so forth?
     
    Last edited: Oct 10, 2006
  12. Oct 10, 2006 #11

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    If it helps you, sure.
     
  13. Oct 11, 2006 #12

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Only if afterwards you plug the (k+x)'s back in, so you can see what your new thing looks like.

    And I disagree, the problem you posted does deal with factorials.


    Just to confirm, you did figure out how the first problem became (n+2)! - 1 right?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help on factorials
  1. Factorial of infinity (Replies: 1)

  2. Factorials of Fractions (Replies: 16)

  3. Inverse Factorials (Replies: 13)

Loading...