TITLE: $\sum_{k=0}^{3^n-1}\binom{2k}{k}$ is divisible by $3^n$
QUESTION [9 upvotes]: Prove that the sum $$\sum_{k=0}^{3^n-1}\binom{2k}{k}$$ is divisible by $3^n$.
 Use by Vandermonde's Identity. So
$$\binom{2k}{k}
=\sum_{j=0}^{k}\binom{k}{j}^2 $$.
Any hints or solutions are greatly appreciated.

REPLY [3 votes]: Since $\binom{2k}{k}=[x^0](x+x^{-1})^{2k}$ we have that
$$ \sum_{k=0}^{3^n-1}\binom{2k}{k} = [x^0]\sum_{k=0}^{3^n-1}(x+x^{-1})^k = [x^0]\frac{(x+x^{-1})^{3^n}-1}{(x+x^{-1})-1}=\text{Res}\left(\frac{(z+z^{-1})^{3^n}-1}{z^2-z+1},z=0\right)$$
but actually $\text{Res}\left(\frac{z^m}{z^2-z+1},z=0\right)$ has a pretty simple behaviour. It equals zero for any non-negative integer $m$ and for every $m\in3\mathbb{Z}$, $1$ for every negative $m\in 6\mathbb{Z}+\{4,5\}$ and $-1$ for every negative $m\in 6\mathbb{Z}+\{1,2\}$. You may draw interesting conclusions from expanding $(z+z^{-1})^{3^n}$ through the binomial theorem and considering that $\binom{3^n}{h}$, for any non-negative $h\not\in 3\mathbb{Z}$, is divided by a large power of $3$.