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Math Help - Sum of binomial coefficients

  1. #1
    Senior Member
    Joined
    Nov 2007
    Posts
    329

    Sum of binomial coefficients

    \left( {\begin{array}{*{20}c}<br />
   {2008}  \\<br />
   0  \\<br />
\end{array}} \right) + \left( {\begin{array}{*{20}c}<br />
   {2008}  \\<br />
   4  \\<br />
\end{array}} \right) + \left( {\begin{array}{*{20}c}<br />
   {2008}  \\<br />
   8  \\<br />
\end{array}} \right) + ... + \left( {\begin{array}{*{20}c}<br />
   {2008}  \\<br />
   {2008}  \\<br />
\end{array}} \right)=?
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  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Let F(x)=(1+x)^{2008} and \zeta=e^{\frac{2\pi\cdot{i}}{4}}

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

    Then: 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; April 3rd 2008 at 05:42 AM.
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