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Math Help - Simplify a factorial

  1. #1
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    Question Simplify a factorial

    Hello! Thanks for taking the time to look at my post. I am studying sequences and series and I don't understand the part about factorials. An example problem is below:

    [(2n-1)!]/[(2n+1)!]

    And another:

    [(2n+2)!]/[(2n)!]

    My goal for both of the above is to simplify the ratio of factorials. I know that a factorial is the product of all the numbers (i.e. 6! is 1*2*3*4*5*6= 720) but I can't grasp how that works when the n's get thrown in there. Any help would be greatly appreciated!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UDaytonFlyer View Post
    Hello! Thanks for taking the time to look at my post. I am studying sequences and series and I don't understand the part about factorials. An example problem is below:

    [(2n-1)!]/[(2n+1)!]
    hint: (2n + 1)! = (2n + 1)(2n)(2n - 1)(2n - 2) \cdots 2 \cdot 1

    (2n+2)!]/[(2n)!]
    hint: (2n + 2)! = (2n + 2)(2n + 1)(2n) \cdots 2 \cdot 1


    look at the tail of both factorials that i gave you. what are they?
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  3. #3
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    Quote Originally Posted by UDaytonFlyer View Post
    Hello! Thanks for taking the time to look at my post. I am studying sequences and series and I don't understand the part about factorials. An example problem is below:

    [(2n-1)!]/[(2n+1)!]

    And another:

    [(2n+2)!]/[(2n)!]

    My goal for both of the above is to simplify the ratio of factorials. I know that a factorial is the product of all the numbers (i.e. 6! is 1*2*3*4*5*6= 720) but I can't grasp how that works when the n's get thrown in there. Any help would be greatly appreciated!
    (2n+1)! = (2n+1)\times 2n \times (2n-1)!

    So \frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n+1)\times 2n \times (2n-1)!} = \frac{1}{2n(2n+1)}
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    Quote Originally Posted by Jhevon View Post
    hint: (2n + 1)! = (2n + 1)(2n)(2n - 1)(2n - 2) \cdots 2 \cdot 1

    hint: (2n + 2)! = (2n + 2)(2n + 1)(2n) \cdots 2 \cdot 1


    look at the tail of both factorials that i gave you. what are they?
    Hm, I think it would be the last two "terms" of the factorial? and they would "cancel" when divided by a similar factorial because they share the same terms?
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    Quote Originally Posted by Prove It View Post
    (2n+1)! = (2n+1)\times 2n \times (2n-1)!

    So \frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n+1)\times 2n \times (2n-1)!} = \frac{1}{2n(2n+1)}
    So for the (2n-1) part you stopped expanding the factorial there because from that point forward the two factorials would have the same terms and thus cancel, right?
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  6. #6
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    Quote Originally Posted by UDaytonFlyer View Post
    So for the (2n-1) part you stopped expanding the factorial there because from that point forward the two factorials would have the same terms and thus cancel, right?
    That's right
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