<s>[INST] "Ballot numbers" sum up to Catalan numbers 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 $n$.
 [/INST] Notice that:
$$
   \sum_{k=1}^n \frac{k^2}{n} \binom{2n-k-1}{n-1} \stackrel{k \to n+1-k}{=} \sum_{k=1}^n \frac{(n+1-k)^2}{n} \binom{n+k-2}{n-1} \stackrel{\binom{a}{b} = \binom{a}{a-b}}{=} \sum_{k=1}^n \frac{(n+1-k)^2}{n} \binom{n+k-2}{k-1} = \sum_{k=0}^{n} \frac{(n-k)^2}{n} \binom{n+k-1}{k}
$$
This this nothing but a convolution of two sequences $a_{k} = \frac{k^2}{n}$ and $b_{k} = \binom{n+k-1}{k}$. Therefore the generating function for the sum $c_n = \sum_{k=0}^{n} a_{n-k} b_k$ is the product of generating functions of $a_k$ and $b_k$:
$$
   \sum_{k=0}^\infty x^k a_k = \frac{1}{n} \sum_{k=0}^\infty x^k k^2 = \frac{1}{n} \left( x\frac{\mathrm{d}}{\mathrm{d} x}\right)^2 \sum_{k=0}^\infty x^k = \frac{1}{n} \frac{x(1+x)}{(1-x)^3}
$$
$$
   \sum_{k=0}^\infty \binom{n+k-1}{k} x^k = \frac{1}{(1-x)^n}
$$
Thus:
$$
    c_n = [x]^n \left( \frac{1}{n} \frac{x(1+x)}{(1-x)^3}  \frac{1}{(1-x)^n} \right) = [x]^n \left( \frac{1}{n} \frac{x}{(1-x)^{n+3}} + \frac{1}{n} \frac{x^2}{(1-x)^{n+3}} \right) = \frac{1}{n} \left(\binom{2n+1}{n-1} + \binom{2n}{n-2}  \right)
$$
Now, using 
$$
   \binom{2n}{n-2} = \binom{2n+1}{n-1} - \frac{2n}{n-1} \binom{2n-1}{n-2}
$$
which is easily proven using recurrence relation for factorials comprising binomial coefficients,  we get
$$
   c_n = \frac{2}{n} \binom{2n+1}{n-1} - \frac{2}{n-1} \binom{2n-1}{n-2} = C_{n+1} - C_{n}
$$
</s>