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Math Help - factorial notation help

  1. #1
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    factorial notation help

    (n+2)/n!=56

    I'm new to this and having trouble trouble breaking this down, could someone please give me a hint to get this started?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    you are saying to divide  n+2 by  n! and get 56.

    This can never happen though...
    Are you sure you have the right problem?
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  3. #3
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    Sorry,

    (n+2)!/n!=56
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Note that  (n+2)! = (n+2)(n+1)(n)(n-1)(n-2) \dots = (n+2)(n+1)n!

    So  \frac{(n+2)!}{n!} = \frac{(n+2)(n+1)n!}{n!}

    Do you see where to go now?
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  5. #5
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    Given: (n + 2)!/n! = 56
    (n + 2)(n + 2 - 1)(n + 2 - 2)!/n! = 56
    (n + 2)(n + 1) = 56
    n + 2 = 56 or n + 1 = 56
    Which gives n = 54, 55
    May be it helps You
    Last edited by mr fantastic; December 9th 2008 at 12:54 AM. Reason: Deleted link to commercial website
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    You solved that wrong...

     (n+2)(n+1) = n^2+3n+2=56

    So  n^2+3n-54 = 0

     (n+9)(n-6) = 0

     n=-6,9
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  7. #7
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    thanks for the help chip, when down to

    (n+9)(n-6) how do you know which to make positive and negative?

    eg how do you know its not (n+6)(n-9) without substiting into the initial equation?
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    its simple factoring

    given that  (n-a)(n-b) = n^2+3n-54

    doing foil...  n^2+(-a-b)n+ab = n^2+3n-54

    So  a+b = -3 and  ab = -54

    By inspection it can be seen  a=-9 , \; b=6
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  9. #9
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    Quote Originally Posted by chiph588@ View Post
    You solved that wrong...

     (n+2)(n+1) = n^2+3n+2=56

    So  n^2+3n-54 = 0

     (n+9)(n-6) = 0

     n=-6,9
    Small correction: n = -9, 6.

    But only n = 6 is a valid solution to the original equation (to the OP: why?).
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