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Solve (2n-1)!

  1. Oct 6, 2008 #1
    Can someone please explain to me why (2n-1)! = (2n-1)(2n)(2n-1)! ?? I'm very confused.
     
  2. jcsd
  3. Oct 6, 2008 #2

    marcusl

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    Re: (2n-1)!

    It doesn't. Did this come from a book?
     
  4. Oct 7, 2008 #3

    gabbagabbahey

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    Re: (2n-1)!

    Do you mean (2n)!= (2n)(2n-1)! ? If so, it comes from the definition of factorial: (2n)!=(2n)(2n-1)(2n-2)....(1)
    and (2n-1)!=(2n-1)(2n-2)....(1) so (2n)(2n-1)!=(2n)(2n-1)(2n-2)...(1)=(2n)!
     
  5. Oct 7, 2008 #4

    HallsofIvy

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    Re: (2n-1)!

    Yes, you are! Dividing both sides of your formula by (2n-1)! you get 1= (2n-1)(2n) which is NOT true!

    Perhaps you mean (2n+1)!= (2n+1)(2n)(2n-1)!. That's true because, by definition, (2n+1)!= (2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)...(3)(2)(1). And (2n-1)!= (2n-1)(2n-3)(2n-4)...(3)(2)(1), the "tail end" of that first product. so (2n+1)!= (2n+1)(2n)(2n-1)!
     
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