TITLE: Binomial Identity
QUESTION [8 upvotes]: I recently noted that 
$$\sum_{k=0}^{n/2} \left(-\frac{1}{3}\right)^k\binom{n+k}{k}\binom{2n+1-k}{n+1+k}=3^n$$
Is this a known binomial identity? Any proof or reference?

REPLY [3 votes]: This is a response to Joe Alfano's comment. Define the two discrete functions
$$F(n,k):=(-1)^k3^{n-k}\binom{n+k}k\binom{2n-k-2}{n-2k} \qquad \text{and} \qquad
G(n,k):=(-1)^{k-1}3^{n+1-k}\binom{n+k-1}{k-1}\binom{2n-k-1}{n-2k}.$$
The proof for the identity $\sum_{k\geq0}F(n,k)=0$ follows from
$$F(n,k)=G(n,k+1)-G(n,k).\tag1$$
Equation (1) can be verified say up on dividing through by $F(n,k)$ and simplifying. The rest comes from summing (1) over all integers $k$ and where the convention $\binom{a}b=0$ (if $b>a$ or $b<0$) is brought to bear. In this regard, the RHS of (1) telescopes $\sum_kG(n,k+1)-\sum_kG(n,k)=0$. Hence $\sum_{k\in\mathbb{Z}}F(n,k)=\sum_{k=0}^{\lfloor\frac{n}2\rfloor}F(n,k)=0$, as desired.
=================
Now, back to the original problem from the OP. Define the functions
$$A(n,k):=\frac{(-1)^k}{3^{n+k}}\binom{n+k}k\binom{2n+1-k}{n-2k}, \qquad \text{and} \qquad
B(n,k):=\frac{(-1)^{k-1}}{3^{n+k}}\binom{n+k}{k-1}\binom{2n+2-k}{n+1-2k}.$$
Again, one checks $A(n+1,k)-A(n,k)=B(n,k+1)-B(n,k)$ and then sum over the integers $k$. The outcome is $\sum_kA(n+1,k)-\sum_kA(n,k)=0$. That is to say,
the sum $\sum_kF(n,k)=\sum_{k=0}^{n/2}\frac{(-1)^k}{3^{n+k}}\binom{n+k}k\binom{2n+1-k}{n-2k}$ is identically a constant. A quick check at $n=0$ reveal that the constant is $1$. The proof ends here.