I want to find(?) derive(?) obtain(?) closed formula for
$$\sum_{k=0}^{n+1} \sum_{r=0}^{k-1}\binom{ n+1}{k}\binom{n}{r}$$
Actually for some integers I found this as $2^{2n}$ ....
My first trial is using double sum technique in $\sum_{k=0}^n \sum_{l=0}^k \binom{n}{k} \binom{k}{l} (-1)^{k-l} s_l ?= \sum_{l=0}^n \sum_{k=l}^n (-1)^{n-k} \binom{n}{k}\binom{k}{l}s_l $
and then via mathematica by some $n$,
\begin{align} \sum_{k=0}^{n+1} \sum_{r=0}^{k-1} \begin{pmatrix} n+1 \\ k \end{pmatrix}\begin{pmatrix} n \\ r \end{pmatrix} = \sum_{r=0}^{n+1} \sum_{k=r+1}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} \begin{pmatrix} n\\ r \end{pmatrix} \end{align} But still does not have explicit form...
