Hard elementary combinatorics problem How does one compute (without brute force) the smallest integer $n$ such that
$\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?

Here is a hint. Prove the identity
\begin{align}
\sum_{i = 1}^{n} \binom{2n}{2i-1} x^{i-1} = \frac{(1 + 2 \sqrt{x} + x)^{n} - (1 - 2 \sqrt{x} + x)^{n} }{2 \sqrt{x}}.
\end{align}
Setting $x = -3$, we have
\begin{align}
\frac{(1 + 2 \sqrt{-3} -3)^{n} - (1 - 2 \sqrt{-3} -3)^{n} }{2 \sqrt{-3}} = \frac{4^{n}}{\sqrt{3}} \sin (\tfrac{2 \pi n}{3})
\end{align}
Using an argument involving periodicity, you should conclude that this sum vanishes infinitely often and you can therefore determine the smallest $n$ accordingly, which is $n = 3$.
