There is a nice proof on MSE which shows that for $n \ge 1$, $$ \sum_{k = 0}^{n}{4n \choose 4k} = {1 \over 4}\left[\left(-1\right)^{n}\, 2^{2n + 1} + 16^{n}\right] $$
I wonder if there is a more combinatorial proof for this?
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2Yes! See Third and Fourth Binomial Coefficients by Arthur Benjamin and Jacob Scott, available online at https://scholarship.claremont.edu/hmc_fac_pub/124/. – Mike Earnest Feb 07 '19 at 17:27
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When $n = 0$, LHS $= 1$ while RHS $= 3/4$ !!!. – Felix Marin Feb 07 '19 at 21:32
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@FelixMarin It is correct for all $n\ge 1$. To make it correct for $n=0$, change the RHS to $\frac14[2(-4)^n+16^n+0^n]$. – Mike Earnest Feb 07 '19 at 22:25
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$\displaystyle\sum_{k = 0}^{n}{4n \choose 4k} = {1 \over 4}\left[\left(-1\right)^{n}, 2^{2n + 1} + 16^{n} + \delta_{n0}\right] $ – Felix Marin Feb 07 '19 at 22:29