Catalan numbers, Condition that $\sum_{k=1}^n c_k=1 \mod 3$ using Lucas theorem Catalan numbers are $c_n=\frac{1}{n+1}{\binom {2n}{n}}$.

Prove that $\sum_{k=1}^n c_k\equiv 1 \mod 3 \iff$ The digit $2$ appears in the ternary representation of $n+1$.

I was shown the solution a while ago and forgot it: It relies on finding some short identity that $c_n$ satisfy when considered modulo $3$, and using Lucas' theorem:
$\binom{n}{k}$ is divisible by 3 iff when adding $k+(n-k)$ in base $3$ there is a carry.

I was told the main part of the proof: we show that $\sum_{k=1}^n c_k -1 \equiv \binom{2(n+1)}{n+1} \mod 3$, and then we get $$\sum_{k=1}^n c_k -1 \equiv 0 \mod 3 \iff      \binom{2(n+1)}{n+1} \equiv 0 \mod 3$$
$ \iff$there is a carry when adding $(n+1)+(n+1)$ in base $3$ $\iff$ The digit $2$ appears in the ternary representation of $n+1$.
Therefore, it suffices to prove $c_k \equiv \binom{2k+2}{k+1} - \binom{2k}{k} \mod 3$, and we will get by a telescopic series what we wanted. Can someone now tie the last knot and prove the congruence?
