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I'm going through my workbook and ran across this: $$\sum_{k=0}^{n-1}\binom{4n}{4k+1}=\frac{1}{2}\sum_{k=0}^{n-1}\binom{4n}{4k+1}+\frac{1}{2}\sum_{k=0}^{n-1}\binom{4n}{4(n-k-1)-1}$$ $$=\frac{1}{2}\sum_{k=0}^{2n-1}\binom{4n}{2k+1}$$ The first thing that bothers me is why does: $$\binom{4n}{4k+1}=\binom{4n}{4(n-k-1)-1}$$ The second thing that bothers me is why the sum of those equals: $$\frac{1}{2}\sum_{k=0}^{2n-1}\binom{4n}{2k+1}$$

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  • $\begingroup$ The fist is just a relabeling $k-> n-k$ $\endgroup$ – tired Nov 6 '15 at 16:44
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I think that ${4n\choose4(n-k-1)-1}$ should be ${4n\choose4(n-k)-1}$. That is the basic symmetry of binomial coefficients, that ${4n\choose x}={4n\choose 4n-x}$.
The second thing: the first sum includes all odd numbers that are 1 more than a multiple of 4; the second sum includes all odd numbers that are 1 less than a multiple of 4. Together, they include all odd numbers below $4n$.

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