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Math Help - Prove that...(using Binomial theorem)?

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

    Prove that:
    {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
    = \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: S(x)=<br />
\sum\limits_{k = 1}^n {\binom{n}{k} \cdot \tfrac{{\left( {1 - x} \right)^k }}<br />
{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} } <br />

    Exchange the integral and the sum: <br />
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} <br />
 = \int_0^1 {\tfrac{1}<br />
{z} \cdot \left[ {1 - \left( {z \cdot x - z + 1} \right)^n } \right]dz} <br /> <br />

    Consider the change of variables: <br />
{z \mapsto z \cdot x - z + 1}<br />
to get: <br />
S\left( x \right) = \int_x^1 {\tfrac{{1 - z^n }}<br />
{{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}}<br />
{1} + ... + \tfrac{{1 - x^n }}<br />
{n}<br />
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