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

Proof
Let:
 * $\displaystyle S := \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}$

Then:

Also:

Reassembling $S$:


 * $\displaystyle S = \sum_k \binom {1 + 2 k} k \binom {- m - 2 k - 1} {n - m - k} \frac {\left({-1}\right)^{n - m} } {1 + 2 k}$

Recall Sum over $k$ of $\dbinom {r - t k} k$ by $\dbinom {s - t \left({n - k}\right)} {n - k}$ by $\dfrac r {r - t k}$:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

Making the following substitutions:
 * $r \gets 1$
 * $t \gets -2$
 * $n \gets n - m$

we obtain:

Thus: