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Math Help - Binomial Theorem?

  1. #1
    Super Member fardeen_gen's Avatar
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    Binomial Theorem?

    Prove that:

    (n!m!)/(m + n)! = (nC1)/(m + 1) - 2.(nC2)/(m + 2) + 3.(nC3)/(m + 3) - ... + (-1)^(n+1)n.(nCn)/(m + n)

    NOTE: nC1 means n choose 1. I struggled with latex so I had to eventually use this notation.
    Last edited by fardeen_gen; February 9th 2009 at 10:28 AM.
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  2. #2
    Super Member fardeen_gen's Avatar
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    This question seems similar:

    Find a closed form for the expression

    for .












    .

    But I am unable to form the required integral in my problem.
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  3. #3
    MHF Contributor
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    Yes you are right this is the same kind of problem

    -1)^{k-1}}{m+k} = \sum_{k=1}^{n} \binom{n}{k} \int_{0}^{1} k\-1)^{k-1} x^{m+k-1} dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \sum_{k=1}^{n} \binom{n}{k} \int_{0}^{1} k\-1)^{k-1} x^{m+k-1} dx" />

    -1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-1)^{k-1} x^{k-1} dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-1)^{k-1} x^{k-1} dx" />

    -1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-x)^{k-1} dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-x)^{k-1} dx" />

    Now
    \sum_{k=0}^{n} \binom{n}{k} (-x)^{k} = (1-x)^n

    By differentiation with respect to x
    -x)^{k-1} = n\1-x)^{n-1}" alt="\sum_{k=1}^{n} \binom{n}{k} k\-x)^{k-1} = n\1-x)^{n-1}" />

    Therefore
    -1)^{k-1}}{m+k} = \int_{0}^{1} x^m \:n\1-x)^{n-1}dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \:n\1-x)^{n-1}dx" />

    -1)^{k-1}}{m+k} = n\:\int_{0}^{1} x^m \1-x)^{n-1}dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = n\:\int_{0}^{1} x^m \1-x)^{n-1}dx" />

    Now you just need to show that
    1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}" alt="n\:\int_{0}^{1} x^m \1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}" />
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