The $n$-th Catalan number is defined as $C_n:=\frac{1}{n+1}\binom{2n}{n}=\frac{1}{n}\binom{2n}{n+1}$.

I have found the following two identities involving Catalan numbers, and my question is if anybody knows them, or if they are special cases of more general results (references?):

(1) For any $n\geq 1$ we have \begin{equation} \binom{2n}{n} + \binom{2n}{n-1} = \sum_{i=0}^{n-1} (4i+3)C_i C_{n-i-1} \enspace. \quad (1) \end{equation}

(2) For any $n\geq 1$ and $k=n,n+1,\ldots,2n-1$ we have \begin{equation} \frac{k-n+1}{n}\binom{2n}{k+1}=\sum_{i=0}^{2n-1-k} \frac{2k-2n+1}{k-i}\binom{2(n-i-1)}{k-i-1} \cdot C_i \enspace. \quad (2) \end{equation}

For the special case $k=n$ equation (2) is the well-known relation \begin{equation} C_n=\sum_{i=0}^{n-1} C_{n-1-i} C_i \enspace. \quad (2') \end{equation} For the special case $k=n+1$ equation (2) yields \begin{equation*} C_n=\frac{n+2}{2(n-1)} \sum_{i=0}^{n-2} \frac{3(n-1-i)}{n+1-i} \cdot C_{n-1-i} C_i \enspace, \quad (2'') \end{equation*} a weighted sum with one term less than (2').

I appreciate any hints, pointers etc.!