Help with a question on binomial

Question

Prove that $$\sum_{r=1}^{k}(-3)^{r-1}\dbinom{3n}{2r-1}= 0,$$ where $k=\frac{3n}{2}$, and $n$ is an even positive integer

Answer 1

First, reformulate the question to showing that $\sum_{r\geq0}\binom{6m}{2r+1}(-3)^r=0$ (with $n=2m$); forget the $k$.

Put $\def\ii{\mathbf i}z=1+\sqrt3\ii\in\Bbb C$, and note that $z$ is twice a $6$-th root of unity, so that $z^6=2^6$ is a real number. Then by the binomial formula $$ \sum_{k>0}\binom{6m}k(\sqrt3\ii)^k=z^{6m}=2^{6m}=\bar z\,^{6m}=\sum_{k>0}\binom {6m}k(-\sqrt3\ii)^k. $$ Taking the difference between the two summations, the terms with even $k$ cancel out, and what remains is $$ 0=2\sqrt3\ii\sum_{r\geq0}\binom{6m}{2r+1}(-3)^r, $$ which gives the desired result.

Since $z^3=-8$ is real as well, the same argument shows that you can drop the condition that $n$ is even.