Proving one of the Binomial Identities 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?

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.
