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Math Help - Proof for an alternative binomial coefficient

  1. #1
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    Question Proof for an alternative binomial coefficient

    Could anyone please point me in the direction on how to proof C(-n, k) = (-1)^k C(n + k - 1, k) ?

    I'm having a problem with this as the factorial of a negative integer is infinit.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by chris2547 View Post
    Could anyone please point me in the direction on how to proof C(-n, k) = (-1)^k C(n + k - 1, k) ?

    I'm having a problem with this as the factorial of a negative integer is infinit.

    Thanks in advance!
    You mean C(n,k) instead of C(-n,k)

    Beside (-1)^k is not infinity.
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  3. #3
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    Quote Originally Posted by novice View Post
    You mean C(n,k) instead of C(-n,k)

    Beside (-1)^k is not infinity.
    No, I'm pretty sure I mean C(-n,k); and did you ever try (-n!) ?
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  4. #4
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    Quote Originally Posted by chris2547 View Post
    No, I'm pretty sure I mean C(-n,k); and did you ever try (-n!) ?
    I assume by C(-n,k) you mean the coefficient of x^k in the binomial expansion of (1+x)^{-n}

    CB
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  5. #5
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    Quote Originally Posted by chris2547 View Post
    Could anyone please point me in the direction on how to proof C(-n, k) = (-1)^k C(n + k - 1, k) ?

    I'm having a problem with this as the factorial of a negative integer is infinit.

    Thanks in advance!
    The starting point in this is to generalize the definition of C(x, k) to be

    C(x,k) = \frac{x (x-1) (x-2) \cdots (x-k+1)}{k!}.

    With this definition, x can be any real number, not necessarily an integer. k must be a non-negative integer.

    Now let x = -n and see what you get.
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  6. #6
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    Quote Originally Posted by awkward View Post
    The starting point in this is to generalize the definition of C(x, k) to be

    C(x,k) = \frac{x (x-1) (x-2) \cdots (x-k+1)}{k!}.

    With this definition, x can be any real number, not necessarily an integer. k must be a non-negative integer.

    Now let x = -n and see what you get.
    Thanks alot! I got it now. I just missed that one detail about the generalization of the binomial coŽfficiŽnt.
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