The binomial identity is
$$\sum^{\left \lfloor{n/3}\right \rfloor }_{k=-\left \lfloor{n/3}\right \rfloor} (-1)^k \binom{2n}{n+3k} = 2 \cdot 3^{n-1} $$
and this is valid for all positive integers $n$. What would be some proofs to show that this identity is true?
On
Suppose we seek to prove that
$$\sum_{k=-\lfloor n/3\rfloor}^{\lfloor n/3\rfloor} (-1)^k {2n\choose n+3k} = 2\times 3^{n-1}.$$
We start by introducing the integral
$${2n\choose n+3k} = {2n\choose n-3k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-3k+1}} (1+z)^{2n} \; dz.$$
Observe that this vanishes for $3k\gt n$ (pole canceled) and for $3k\lt -n$ (upper range of polynomial term exceeded) so we may extend the summation to $[-n, n]$ getting
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{2n} \sum_{k=-n}^{n} (-1)^k z^{3k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{2n} (-1)^n z^{-3n} \sum_{k=0}^{2n} (-1)^k z^{3k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{4n+1}} (1+z)^{2n} (-1)^n \frac{1-(-1)^{2n+1} z^{3(2n+1)}}{1+z^3} \; dz .$$
Only the first piece from the difference due to the geometric series contributes and we get
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{4n+1}} (1+z)^{2n} (-1)^n \frac{1}{1+z^3} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{4n+1}} (1+z)^{2n-1} (-1)^n \frac{1}{1-z+z^2} \; dz .$$
We have two poles other than zero and infinity at $\rho$ and $1/\rho$ where
$$\rho = \frac{1+\sqrt{3}i}{2}$$
and using the fact that residues sum to zero we obtain
$$S + \frac{(-1)^n}{\rho(1+\rho)} \frac{1}{\rho-1/\rho} \left(\frac{(1+\rho)^2}{\rho^4}\right)^n + \frac{(-1)^n}{1/\rho(1+1/\rho)} \frac{1}{1/\rho-\rho} \left(\frac{(1+1/\rho)^2}{1/\rho^4}\right)^n \\ + \mathrm{Res}_{z=\infty} \frac{1}{z^{4n+1}} (1+z)^{2n-1} (-1)^n \frac{1}{1-z+z^2} = 0.$$
We get for the residue at infinity
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{4n+1} (1+1/z)^{2n-1} (-1)^n \frac{1}{1-1/z+1/z^2} \\= -\mathrm{Res}_{z=0} z^{2n+2} (1+z)^{2n-1} (-1)^n \frac{1}{z^2-z+1} = 0.$$
Now if $z^2 = z-1$ then $z^4 = z^2-2z+1 = -z$ and thus
$$\frac{(1+1/\rho)^2}{1/\rho^4} = \frac{(1+\rho)^2}{\rho^4} = \frac{\rho-1+2\rho+1}{-\rho} = -3$$
and furthermore with $z(1+z)(z-1/z) = (1+z)(z^2-1)$ and $(1+z)(z-2) = z^2-z-2 = -3$ we finally get
$$S + (-1)^n \times \left(-\frac{1}{3}\right) (-3)^n + (-1)^n \times \left(-\frac{1}{3}\right) (-3)^n = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{S = 2\times 3^{n-1}.}$$
This can be proved by induction on $n$. Instead of writing it up as efficiently as possible, I’ll show you how I got there.
First note that there’s no need to specify limits on the index $k$: the sum is taken over all $k$ for which the binomial coefficient is non-zero, with the usual convention that $\binom{n}k=0$ if $k>n$ or $k<0$. With the induction hypothesis
$$\sum_k(-1)^k\binom{2n}{n+3k}=2\cdot3^{n-1}\;,\tag{1}$$
the natural approach to the induction step looks like this:
$$\begin{align*} \sum_k(-1)^k\binom{2n+2}{n+1+3k}&=\sum_k(-1)^k\left(\binom{2n}{n-1+3k}+2\binom{2n}{n+3k}+\binom{2n}{n+1+3k}\right)\\ &=4\cdot3^{n-1}+\sum_k(-1)^k\binom{2n}{n-1+3k}+\sum_k(-1)^k\binom{2n}{n+1+3k}\\ &=4\cdot3^{n-1}+\sum_k(-1)^k\binom{2n}{n+1-3k}+\sum_k(-1)^k\binom{2n}{n+1+3k}\\ &=4\cdot3^{n-1}+2\sum_k(-1)^k\binom{2n}{n+1+3k}\;, \end{align*}$$
and we’d like to know that
$$\sum_k(-1)^k\binom{2n}{n+1+3k}=3^{n-1}\;.\tag{2}$$
Fine: we can try proving both identities by induction simultaneously. In that case $(2)$ is also part of our induction hypothesis, alongside $(1)$, so we can finish off the calculation with
$$4\cdot3^{n-1}+2\sum_k(-1)^k\binom{2n}{n+1+3k}=4\cdot3^{n-1}+2\cdot3^{n-1}=2\cdot3^n\;,$$
as desired, and the rest of our induction step looks like this:
$$\begin{align*} \sum_k(-1)^k\binom{2n+2}{n+2+3k}&=\sum_k(-1)^k\left(\binom{2n}{n+3k}+2\binom{2n}{n+1+3k}+\binom{2n}{n+2+3k}\right)\\ &=2\cdot3^{n-1}+2\cdot 3^{n-1}+\sum_k(-1)^k\binom{2n}{n+2+3k}\\ &=4\cdot3^{n-1}+\sum_k(-1)^k\binom{2n}{n+2+3k}\\ &=4\cdot3^{n-1}-\sum_k(-1)^k\binom{2n}{n-1+3k}\\ &=4\cdot3^{n-1}-\sum_k(-1)^k\binom{2n}{n+1-3k}\\ &=4\cdot3^{n-1}-\sum_k(-1)^k\binom{2n}{n+1+3k}\\ &=4\cdot3^{n-1}-3^{n-1}\\ &=3^n\;, \end{align*}$$
again as desired.
This is yet another induction in which it seems to be easier to prove a stronger result.