Sum of binomial coefficients

**james_bond** · March 30th 2008, 05:27 AM

$\left( {\begin{array}{*{20}c} {2008} \\ 0 \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} {2008} \\ 4 \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} {2008} \\ 8 \\ \end{array}} \right) + ... + \left( {\begin{array}{*{20}c} {2008} \\ {2008} \\ \end{array}} \right)=?$

**PaulRS** · April 3rd 2008, 05:17 AM

Let $F(x)=(1+x)^{2008}$ and $\zeta=e^{\frac{2\pi\cdot{i}}{4}}$

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

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

Now, calculate the left hand side and divide it by 4 to get the result

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