<s>[INST] calculate $\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}$? prove the following equation :
$$\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}={s+2r-1\choose r} - {s+2r-1\choose r-1}$$
 [/INST] Suppose we seek to verify that
$$\frac{1}{r+1} \sum_{q=0}^r (q+1) {2r-q\choose r-q}
{s-2+q\choose q} = {s+2r-1\choose r} - {s+2r-1\choose r-1}.$$
Introduce the integral representation
$${2r-q\choose r-q}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{(1+w)^{2r-q}}{w^{r-q+1}}
\; dw$$
Note that this is zero when $q\gt r$ so we may extend $q$ to infinity,
getting for the sum
$$\frac{1}{r+1} \frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{(1+w)^{2r}}{w^{r+1}}
\sum_{q\ge 0} (q+1)
{s-2+q\choose q} \frac{w^q}{(1+w)^q}
\; dw.$$
Now observe that
$$z \times \frac{1}{(1-z)^{s-1}}
= \sum_{q\ge 0} {s-2+q\choose q} z^{q+1}$$
and hence
$$\frac{1}{(1-z)^{s-1}} + (s-1) \frac{z}{(1-z)^s}
= \sum_{q\ge 0} (q+1) {s-2+q\choose q} z^q.$$
Evaluating this at $z=w/(1+w)$ so that $1-z = 1/(1+w)$ yields
$$(1+w)^{s-1} + (s-1) w (1+w)^{s-1}.$$
Substitute this into the integral to obtain
$$\frac{1}{r+1} {s+2r-1\choose r}
+ \frac{s-1}{r+1} {s+2r-1\choose r-1}.$$
This is
$$\frac{1}{r+1} {s+2r-1\choose r}
+ \frac{s+r}{r+1} {s+2r-1\choose r-1}
- {s+2r-1\choose r-1}.$$
Now
$$\frac{s+r}{r+1} {s+2r-1\choose r-1}
= \frac{1}{r+1} \frac{(s+2r-1)!}{(s+r-1)! (r-1)!}
\\ = \frac{r}{r+1} \frac{(s+2r-1)!}{(s+r-1)! r!}
= \frac{r}{r+1} {s+2r-1\choose r}.$$
Collecting everything we obtain
$$\frac{1}{r+1} {s+2r-1\choose r}
+ \frac{r}{r+1} {s+2r-1\choose r}
- {s+2r-1\choose r-1}
\\ = {s+2r-1\choose r}
- {s+2r-1\choose r-1}$$
as claimed.
</s>