A binomial identity from Mathematical Reflections Here is the problem:
Let $m,n$ be positive integers with $n>m$. Prove that
$\displaystyle\sum_{k=0}^{\lfloor\frac{n+m}2\rfloor} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{n-1}=\binom{n}{m+1}$
This problem is O243 of Mathematical Reflections. A solution had been published using complex integration (https://www.awesomemath.org/assets/PDFs/MR5sol(1).pdf). However, I would like to see a solution using the difference operator, if any, since the form of the summand brings this to mind.

First of all, it is worth stating explicitly that the problem assumes that $\binom{n}{k}$ is zero when $n < 0$ or $n>k$, even for $k\geqslant 0$. Indeed, otherwise
In[61]:= Table[
 Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1], {k, 0, 
   n}], {n, 1, 5}, {m, 0, n - 1}]

Out[61]= {{0}, {0, 0}, {0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0, 0}}

With those restrictions in place the claimed result is indeed reproduced:
In[65]:= Table[
 Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1] Boole[
    0 <= n - 1 <= n + m - 2 k], {k, 0, n}], {n, 1, 5}, {m, 0, n - 1}]

Out[65]= {{1}, {2, 1}, {3, 3, 1}, {4, 6, 4, 1}, {5, 10, 10, 5, 1}}

In[66]:= Table[Binomial[n, m + 1], {n, 1, 5}, {m, 0, n - 1}]

Out[66]= {{1}, {2, 1}, {3, 3, 1}, {4, 6, 4, 1}, {5, 10, 10, 5, 1}}

With this said, the upper bound of summation over $k$ is $m_\ast = \left\lfloor \frac{m+1}{2} \right\rfloor$:
$$
   \mathcal{S}(n,m)= \sum_{k=0}^{m_\ast} (-1)^k \binom{n}{k} \binom{n+m-2k}{n-1} 
$$
The summand is the hypergeometric term, meaning that 
$$
 r(k) = \frac{c_{k+1}}{c_k} = \frac{(-1)^{k+1} \binom{n}{k+1} \binom{n+m-2(k+1)}{n-1} }{(-1)^k \binom{n}{k} \binom{n+m-2k}{n-1} } = \frac{-n+k}{k+1} \frac{-\frac{m+1}{2} + k}{-\frac{m+n-1}{2} + k} \frac{-\frac{m}{2} + k}{-\frac{m+n}{2}+k}
$$
and therefore
$$
   c_k = c_0 \prod_{q=1}^{k} r(q) = \binom{m+n}{n-1} \frac{(-n)_k}{k!} \frac{\left(-\frac{m}{2}\right)_k \cdot \left(-\frac{m+1}{2}\right)_k}{\left(-\frac{m+n}{2}\right)_k \cdot \left(-\frac{m+n-1}{2}\right)_k}
$$
And thus, since $r(m_\ast) = 0$ we have
$$
   \mathcal{S}(n,m) = \binom{n+m}{n-1} \cdot {}_3F_2\left(\left.\begin{array}{cll} -n & -m/2 & -(m+1)/2 \\ & -(m+n)/2 & -(m+n-1)/2 \end{array} \right| 1 \right)
$$
Now, per this identity:
$$
  {}_3F_2\left(\left.\begin{array}{cll} -n & a & b \\ & d & 1+a+b-d-n \end{array} \right| 1 \right) = \frac{(d-a)_n \cdot (d-b)_n}{(d)_n \cdot (d-a-b)_n}
$$
Using this identity for $a = -m/2$, $b = -(m+1)/2$ and $d=\epsilon-(m+n)/2$ with the intent to consider the limit of $\epsilon \to 0$ we have
$$
   \mathcal{S}(n,m) = \binom{n+m}{n-1} \lim_{\epsilon \to 0} \frac{ \left(-\frac{n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n}{2} + \epsilon \right)_n}{ \left(-\frac{m+n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n-m}{2} + \epsilon \right)_n }
$$
Using
$$
   \left(-\frac{n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n}{2} + \epsilon \right)_n = \frac{1}{2^{2n}} \frac{\Gamma(n+2 \epsilon)}{\Gamma(-n+2\epsilon)} = (-1)^n \frac{\Gamma(1+n-2\epsilon) \Gamma(n+2\epsilon)}{2^{2n} \pi} \sin(2 \pi \epsilon) 
$$
$$
 \left(-\frac{m+n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n-m}{2} + \epsilon \right)_n = \frac{\Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) }{\Gamma\left(\frac{-n-m}{2} + \epsilon\right)\Gamma\left(\frac{1+m-n}{2} + \epsilon\right)} = \frac{\Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) }{-\frac{2\pi^2}{\sin(\pi m) + \sin(\pi n - 2 \pi \epsilon)}} \Gamma\left(1 + \frac{m+n}{2} - \epsilon\right) \Gamma\left( \frac{1+n-m}{2} - \epsilon\right) = (-1)^n \frac{\sin(2\pi \epsilon)}{2 \pi^2} \Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) \Gamma\left(1 + \frac{m+n}{2} - \epsilon\right) \Gamma\left( \frac{1+n-m}{2} - \epsilon\right)
$$
Combining, and using the duplication formula we get
$$
    \mathcal{S}(n,m) = \binom{n+m}{n-1} \frac{n! (n-1)!}{(n+m)! (n-m-1)!} = \binom{n}{m+1}
$$
