An interesting combinatorial relation

[texdisplay]\left( \begin{array}{c} n \\ 2k+1 \end{array} \right) = \sum_{i=k+1}^{n-k} \left( \begin{array}{c} i-1 \\ k \end{array} \right) \left( \begin{array}{c} n-i \\ k \end{array} \right)[/texdisplay]

The proof is left as an exercise.