# Sum of binomial coefficients

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• March 30th 2008, 05:27 AM
james_bond
Sum of binomial coefficients
$\left( {\begin{array}{*{20}c}
{2008} \\
0 \\
\end{array}} \right) + \left( {\begin{array}{*{20}c}
{2008} \\
4 \\
\end{array}} \right) + \left( {\begin{array}{*{20}c}
{2008} \\
8 \\
\end{array}} \right) + ... + \left( {\begin{array}{*{20}c}
{2008} \\
{2008} \\
\end{array}} \right)=?$
• April 3rd 2008, 05:17 AM
PaulRS
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