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Math Help - Binomial Formula

  1. #1
    Junior Member
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    Binomial Formula

    Prove, that for every integer n>1,

    \binom{n}{1} - 2 \binom{n}{2} +3 \binom{n}{3} +...+ (-1)^{n-1}n \binom{n}{n}=0

    I get \sum_{x=0}^{n} (-1)^{x-1}x \binom{n}{x}

    The teacher gave us a hint that said we should differentiate and then plug in x =-1. I'm having a hard time converting this to the binomial theorem and understand why x=-1 shows it for all 'n'
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  2. #2
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    (1+x)^n = 1 + \binom{n}{1} x + \binom{n}{2} x^2 + \binom{n}{3} x^3 + \binom{n}{4} x^4 ..... \binom{n}{n} x^n


    differentiate both sides with respect to x

    n(1+x)^{n-1} = \binom{n}{1}  + 2 \binom{n}{2} x +3 \binom{n}{3} x^2 + 4 \binom{n}{4} x^3 ..... n \binom{n}{n} x^{n-1}

    sub x = -1

    n(1-1)^{n-1} = \binom{n}{1}  - 2 \binom{n}{2} +3 \binom{n}{3} - 4 \binom{n}{4} ..... + (-1)^{n-1} n \binom{n}{n}


    0 = \binom{n}{1}  - 2 \binom{n}{2} +3 \binom{n}{3} - 4 \binom{n}{4} ..... + (-1)^{n-1} n \binom{n}{n}
    Last edited by bobak; February 20th 2008 at 01:41 AM. Reason: fixed typo, Thank you James Bond
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  3. #3
    Senior Member
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    Just a little typo: (-1)^{n-1} twice...
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