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Thread: [SOLVED] Using Binomial coefficients in expansion of (1 + x)^n, prove that...?

  1. April 30th 2009, 08:08 PM #1
    fardeen_gen
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    [SOLVED] Using Binomial coefficients in expansion of (1 + x)^n, prove that...?

    If {n\choose 0}, {n\choose 1}, {n\choose 2}, ... , {n\choose n} denote the binomial coefficients in the expansion of (1 + x)^n, prove that \sum_{r = 0}^{n} (-1)^r {n\choose r}\frac{1 + r\log_{e} 10}{(1 + \log_{e} 10^n)^r} = 0.
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  2. June 3rd 2009, 05:14 PM #2
    the_doc
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    The way I did it:

    Spoiler:


    Considering the sum

    \sum_{r = 0}^{n} (-1)^r {n\choose r} (1 + r\ln 10) x^r<br />

    which is equal to the given sum when

    x = \frac{1}{1+ n \ln 10}

    we have

    \sum_{r = 0}^{n} (-1)^r {n\choose r} (1 + r\ln 10) x^r<br /> = \sum_{r = 0}^{n} (-1)^r {n\choose r} x^r + \ln 10 \, \sum_{r = 0}^{n} (-1)^r {n\choose r} r x^r,

    = (1-x)^n + \ln 10 \sum_{r = 0}^{n} (-1)^r {n\choose r} r x^r,

    = (1-x)^n + \ln 10 \, \, x \frac{d}{dx} \left( \sum_{r = 0}^{n} (-1)^r {n\choose r} x^r \right),

    = (1-x)^n + \ln 10 \, \, x \frac{d}{dx} \left( (1-x)^n \right),

    = (1-x)^n - n x (1-x)^{n-1} \ln 10,

    = (1-x)^{n-1} \left( 1-x - n x \ln 10 \right) ,

    = (1-x)^{n-1} \left( 1- (1+ n \ln 10)x \right) .

    Hence

    \sum_{r = 0}^{n} (-1)^r {n\choose r} \frac{1 + r\ln 10}{(1+n \ln 10)^r} =<br /> \left(1-\frac{1}{1+ n \ln 10} \right)^{n-1} \left( 1- (1+ n \ln 10) \cdot \frac{1}{1+ n \ln 10} \right)<br /> ,

    = 0.

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