Identity involving difference of binomial coefficients I am trying to prove the following identity but not sure how to prove it.
[The followings are equivalent forms of the original equality I asked.]
$$
\binom{m+n}{s+1} - \binom{n}{s+1} = \sum_{i=0}^s \frac{m}{s-i+1}\binom{m+1+2(s-i)}{s-i}\binom{n-2(s-i+1)}{i}.
$$
$$
{m+n\choose s+1} - {n\choose s+1}
= \sum_{q=0}^s \frac{m}{q+1} 
{m+1+2q\choose q} {n-2-2q\choose s-q}
$$
$$
\binom{m+n}{s} = \sum_{i=0}^{s} \frac{m}{m+2i}
\binom{m+2i}{i}\binom{n-2i}{s-i}
$$
[Please ignore my attempt. It only explains one of equivalent forms.]
My attempt is to use the combinatorial argument.
The lefthand side could be understood as follow. 
Suppose we have a box containing $m$ black balls and $n$ white balls.
We randomly draw $s+1$ balls out of it.
Then the LHS represents the number of ways that the selected $s+1$ balls.
However, not sure how to make a combinatorial argument on the RHS.
Based on the RHS, the sum of all cases of drawing $s-i$ balls from somewhere and $i$ balls from white balls.
But it is unclear to me to show the above identity.
Any suggestions/answers would be very appreciated. Thanks.

Remark. What follows answers one  of several queries that appeared
at this post, which each query replacing the previous one. We  suggest
making a list so that all the different varieties may be examined.
Starting from the claim
$$\bbox[5px,border:2px solid #00A000]{
{m+n\choose s+1} - {n\choose s+1}
= \sum_{q=0}^s \frac{m}{q+1} 
{m+1+2q\choose q} {n-2-2q\choose s-q}}$$
we observe that
$${m+1+2q\choose q+1} - {m+1+2q\choose q}
\\ = \frac{m+1+q}{q+1} {m+1+2q\choose q} - {m+1+2q\choose q}
\\ = \frac{m}{q+1} {m+1+2q\choose q}.$$
Therefore we have two sums,
$$\sum_{q=0}^s
{m+1+2q\choose q+1} {n-2-2q\choose s-q}
- \sum_{q=0}^s
{m+1+2q\choose q} {n-2-2q\choose s-q}.$$
For the first one we write
$$\sum_{q=0}^s
[w^{q+1}] (1+w)^{m+1+2q} [z^{s-q}] (1+z)^{n-2-2q}
\\ = \underset{w}{\mathrm{res}}\; 
 (1+w)^{m+1} [z^s] (1+z)^{n-2}
\sum_{q=0}^s \frac{1}{w^{q+2}} z^q (1+w)^{2q} (1+z)^{-2q}.$$
We  may extend  $q$ beyond  $s$ because  of the  coefficient extractor
$[z^s]$ in front, getting
$$\underset{w}{\mathrm{res}}\; \frac{1}{w^2} (1+w)^{m+1} [z^s] (1+z)^{n-2}
\sum_{q\ge 0} z^q w^{-q} (1+w)^{2q} (1+z)^{-2q}
\\ =\underset{w}{\mathrm{res}}\; (1+w)^{m+1} [z^s] (1+z)^{n-2}
\frac{1}{w^2} \frac{1}{1-z(1+w)^2/w/(1+z)^2}
\\ =\underset{w}{\mathrm{res}}\; (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{w} \frac{1}{w(1+z)^2-z(1+w)^2}.$$
Repeat the calculation for the second one to get
$$\underset{w}{\mathrm{res}}\;  (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{w(1+z)^2-z(1+w)^2}.$$
Now we have
$$\left(\frac{1}{w}-1\right)\frac{1}{w(1+z)^2-z(1+w)^2}
= \frac{1}{w-z} \frac{1}{w(1+w)}
- \frac{1}{1-wz} \frac{1}{1+w}
\\ = \frac{1}{1-z/w} \frac{1}{w^2(1+w)}
- \frac{1}{1-wz} \frac{1}{1+w}.$$
We thus obtain two components, the first is
$$\underset{w}{\mathrm{res}}\;  (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{1-z/w} \frac{1}{w^2(1+w)}
\\ = \underset{w}{\mathrm{res}}\; \frac{1}{w^2} (1+w)^{m} 
[z^s] (1+z)^{n} \frac{1}{1-z/w} 
\\ = \underset{w}{\mathrm{res}}\; \frac{1}{w^2} (1+w)^{m} 
\sum_{q=0}^s {n\choose q} \frac{1}{w^{s-q}}
= \sum_{q=0}^s {n\choose q}
\underset{w}{\mathrm{res}}\; \frac{1}{w^{s-q+2}} (1+w)^{m} 
\\ = \sum_{q=0}^s {n\choose q} [w^{s-q+1}] (1+w)^m
= [w^{s+1}] (1+w)^m \sum_{q=0}^s {n\choose q} w^q
\\ = - {n\choose s+1} 
+ [w^{s+1}] (1+w)^m \sum_{q=0}^{s+1} {n\choose q} w^q.$$
We may  extend $q$ beyond  $s+1$ due  to the coefficient  extractor in
front, to get
$$- {n\choose s+1} 
+ [w^{s+1}] (1+w)^m \sum_{q\ge 0} {n\choose q} w^q
= - {n\choose s+1} 
+ [w^{s+1}] (1+w)^{m+n}$$
This is
$$\bbox[5px,border:2px solid #00A000]{
{m+n\choose s+1} - {n\choose s+1}.}$$
We have the claim, so we just  need to prove that the second component
will produce zero. We obtain
$$\underset{w}{\mathrm{res}}\;  (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{1-wz} \frac{1}{1+w}
\\ =\underset{w}{\mathrm{res}}\; (1+w)^{m} [z^s] (1+z)^{n}
\frac{1}{1-wz} 
\\ =\underset{w}{\mathrm{res}}\; (1+w)^{m} 
\sum_{q=0}^s {n\choose q} w^{s-q}
= \sum_{q=0}^s {n\choose q} \underset{w}{\mathrm{res}}\; w^{s-q}  (1+w)^{m} 
= 0.$$
This concludes the argument.
