Summing certain numbers and comparing the results with OEIS, I found that $ \sum_{k=1}^n \frac{k^2}{n} \binom{2n-k-1}{n-1} = C_{n+1} - C_{n}, $
where $C_n$ denotes the $n^{\textrm{th}}$ Catalan Number. How can I prove this equation? And is there any combinatorial interpretation?
Some background information: The number $\frac{k}{n} \binom{2n-k-1}{n-1}$ denotes the number of unranked trees of size $n$ with a root degree $k$ (these numbers are known as ballot numbers, see e.g. the book Analytic Combinatorics from Flajolet and Sedgewick, page 68). So one must have $\sum_{k=1}^n \frac{k}{n} \binom{2n-k-1}{n-1} = C_{n}$, since there are $C_n$ many trees of size $n$.