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I'm trying to to combinatorially prove the following identity:

$$\sum_{k=0}^{n} \binom{4n}{4k} = 2^{4n-2} + (-1)^{n}2^{2n-1}$$

However, I don't have any idea how to start or what to count. I would appreciate any suggestions that helps me get the idea how to even start to prove these kind of identities.

Blue
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John
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    Could you please show your progress or any efforts? – Darshan P. Jul 20 '23 at 17:42
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    “how to start” is probably with a much simpler problem. But some things to look for: $\binom{a}{b}$ is the number of subsets of a given $a$-element set that have exactly $b$ element. $2^c$ is the number of subsets of a $c$-element set that have any number of elements. $(-1)^n$ as a factor adds or subtracts depending on whether $n$ is even or odd. – Matthew Leingang Jul 20 '23 at 18:01
  • Here is a different sum involving binomial coefficients with a period of four. The same technique (see my answer) can be used to prove this, but I'm not sure I would call it combinatorial :-) – Jyrki Lahtonen Jul 20 '23 at 18:14

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Expand => $$\frac {(1 + x)^{4n} \pm \left((1 + x)^{2n} + (1 - x)^{2n}\right)}{8}$$

We can substitute x = 1 after equating to the expansion of the above.

Hint: $$\left[\binom{k}{0} + \binom{k}{1} + \dots + \binom{k}{k}\right] = 2^{k}$$

While solving, you will eventually end up at k = 2n, 4n.

Please note: I have added ± so that you can eventually realize whether to subtract or add the binomials when relating $$\binom{4n}{r} \text{ & } \binom{2n}{r - 1}, \binom{2n}{r+1}$$

Hint: You can use Pascal's triangle to see the pattern or something like

$$\sum_{k=0}^n \left[\binom{2n - 1}{2k} + \binom{2n - 1}{2k - 1}\right] = \sum_{k=0}^{n} \binom{2n}{2k}$$

Darshan P.
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