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Thread: Sum of binomial coefficients

  1. #1
    Senior Member
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    329

    Sum of binomial coefficients

    $\displaystyle \left( {\begin{array}{*{20}c}
    {2008} \\
    0 \\
    \end{array}} \right) + \left( {\begin{array}{*{20}c}
    {2008} \\
    4 \\
    \end{array}} \right) + \left( {\begin{array}{*{20}c}
    {2008} \\
    8 \\
    \end{array}} \right) + ... + \left( {\begin{array}{*{20}c}
    {2008} \\
    {2008} \\
    \end{array}} \right)=?$
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  2. #2
    Super Member PaulRS's Avatar
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    Let $\displaystyle F(x)=(1+x)^{2008}$ and $\displaystyle \zeta=e^{\frac{2\pi\cdot{i}}{4}}$

    Note that: $\displaystyle \zeta^{n\cdot{0}}+\zeta^{n\cdot{1}}+\zeta^{n\cdot{ 2}}+\zeta^{n\cdot{3}}=\left\{ \begin{gathered}
    4{\text{ if }} n\equiv{0}(\bmod.4) \hfill \\
    0{\text{ otherwise }} \hfill \\
    \end{gathered} \right.$ ( separate in cases and use the geometric sum when possible)

    Then: $\displaystyle F(\zeta^0)+F(\zeta^1)+F(\zeta^2)+F(\zeta^3)=\sum_{ n=0}^{2008}\binom{2008}{n}\cdot{(\zeta^{n\cdot{0}} +\zeta^{n\cdot{1}}+\zeta^{n\cdot{2}}+\zeta^{n\cdot {3}})}=4\cdot{\sum_{k}{\binom{2008}{4k}}}$

    Now, calculate the left hand side and divide it by 4 to get the result
    Last edited by PaulRS; Apr 3rd 2008 at 05:42 AM.
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