Sum over k of n+k Choose m+2k by 2k Choose k by -1^k over k+1/Proof 1

Proof
From Sum over $k$ of $\dbinom {r - k} m \dbinom {s + k} n$:
 * $\displaystyle \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n = \binom {r + s + 1} {m + n + 1}$

Setting $r = n + k - 1$, $m = 2 k$, $s = 0$, $n = m - 1$ and using $j$ as the index variable, this gives us:
 * $\displaystyle \sum_{j \mathop = 0}^{n + k - 1} \binom {n + k - 1 - j} {2 k} \binom j {m - 1} = \dbinom {n + k} {m + 2 k}$

This leads to:
 * $\displaystyle \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1} = \sum_k \sum_{j \mathop = 0}^{n + k - 1} \binom {n + k - 1 - j} {2 k} \binom {2 k} k \binom j {m - 1} \frac {\left({-1}\right)} {k + 1}$

Note that the terms of the above are $0$ when $n \le j < n + k - 1$.

This means that $k \ge 1$.

So:
 * $0 \le n + k - 1 - j \le k - i < 2 k$

Thus the first binomial coefficient in the above will vanish.

The condition on the second summation can then be replaced by $0 \le j < n$.

After exchange of summation and application of Sum over $k$ of $\dbinom {n + k} {2 k} \dbinom {2 k} k \dfrac {-1^k} {k + 1}$ converts it into:
 * $\displaystyle \sum{_0 \mathop \le j < n} \binom j {m - 1} \delta_{\left({n \mathop - 1 \mathop j}\right) 0}$

where $\delta_{\left({n \mathop - 1 \mathop j}\right) 0}$ denotes the Kronecker delta.

All terms are therefore $0$ except where $j = n - 1$.

Hence the result.