Sum binomial coefficients Show that
$$\sum_{k=0}^{n}\binom{n}{k}\left(\binom{m+k+1}{m-n}+(-1)^{k+1}\binom{m+2(n-k)+1}{m}\right)=0$$
for all integers $m$ and $n$ with $m\ge n\ge 0$.
I tried induction on $n$, but there's not a very nice way to change the LHS from the $n$ case to $n+1$.

We get for the first of the two pieces
$$\sum_{k=0}^n {n\choose k} {m+k+1\choose m-n}
= [z^{m-n}] (1+z)^{m+1} \sum_{k=0}^n {n\choose k} (1+z)^k
\\ = [z^{m-n}] (1+z)^{m+1} (2+z)^n.$$
The second piece is
$$\sum_{k=0}^n {n\choose k} (-1)^{k+1} {m+2(n-k)+1\choose m}
\\ = - [z^m] (1+z)^{m+2n+1}
\sum_{k=0}^n {n\choose k} (-1)^k (1+z)^{-2k}
\\ = - [z^m] (1+z)^{m+2n+1}
\left(1-\frac{1}{(1+z)^2}\right)^n
\\ = - [z^m] (1+z)^{m+1} (2z+z^2)^n
= -[z^m] z^n (1+z)^{m+1} (2+z)^n
\\ = -[z^{m-n}] (1+z)^{m+1} (2+z)^n.$$
This is the claim.
