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Thread: Prove that...(using Binomial theorem)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove that...(using Binomial theorem)?

    Prove that:
    $\displaystyle {n\choose 1}(1 - x) - \frac{{n\choose 2}}{2}(1 - x)^2 + \frac{{n\choose 3}}{3}(1 - x)^3 - ... + (-1)^{n - 1}\cdot \frac{{n\choose n}}{n}(1 - x)^n$
    $\displaystyle = \frac{1 - x}{1} + \frac{1 - x^2}{2} + \frac{1 - x^3}{3} + ... + \frac{1 - x^n}{n}$
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  2. #2
    Super Member PaulRS's Avatar
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    Write: $\displaystyle S(x)=
    \sum\limits_{k = 1}^n {\binom{n}{k} \cdot \tfrac{{\left( {1 - x} \right)^k }}
    {k} \cdot \left( { - 1} \right)^{k + 1} } = \sum\limits_{k = 1}^n { \binom{n}{k} \cdot \left( {1 - x} \right)^k \cdot \left( {\int_0^1 {x^{k - 1} dx} } \right) \cdot \left( { - 1} \right)^{k + 1} }
    $

    Exchange the integral and the sum: $\displaystyle
    S\left( x \right) = \int_0^1 {\left[ {\sum\limits_{k = 1}^n { \binom{n}{k}\cdot z^{k - 1} \cdot \left( {1 - x} \right)^k \cdot \left( { - 1} \right)^{k + 1} } } \right]dz}
    = \int_0^1 {\tfrac{1}
    {z} \cdot \left[ {1 - \left( {z \cdot x - z + 1} \right)^n } \right]dz}

    $

    Consider the change of variables: $\displaystyle
    {z \mapsto z \cdot x - z + 1}
    $ to get: $\displaystyle
    S\left( x \right) = \int_x^1 {\tfrac{{1 - z^n }}
    {{1 - z}}dz} = \int_x^1 {\left( {\sum\limits_{k = 0}^{n - 1} {z^k } } \right)dz} = \sum\limits_{k = 0}^{n - 1} {\left( {\int_x^1 {z^k dz} } \right)} = \tfrac{{1 - x}}
    {1} + ... + \tfrac{{1 - x^n }}
    {n}
    $
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