The even and odd sums can be determined from each other:
$$\sum_{0 \le i < n} \binom{2i}{k} + \sum_{0 \le i < n} \binom{2i+1}{k} = \binom{2n}{k + 1}$$
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$\sum_{0 \le i < n} \binom{2i}{k} + \sum_{0 \le i < n} \binom{2i+1}{k} = \sum_{0 \le i \le 2n-1} \binom{i}{k}$ – Bumblebee Nov 28 '22 at 12:28
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1@Bumblebee Your right-hand side plus the hockey-stick identity yields $\binom{2n}{k+1}$. – Arthur Nov 28 '22 at 12:43
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Does this answer your question? [Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$](https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-zhu-shijie-identity-sum-limits-t-0n-binom-tk) – Bumblebee Nov 30 '22 at 07:53
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@Bumblebee no. The hockey stick identity can be used for a sum over *all* upper indices, but not over only even or only odd indices – cat-sofia Nov 30 '22 at 21:10