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.
 A: 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.
A: This is a mere supplement to @MarkoRiedel's instructive answer slightly streamlining a few steps.

We obtain
\begin{align*}
\color{blue}{\sum_{q=0}^s}&\color{blue}{\frac{m}{q+1}\binom{m+1+2q}{q}\binom{n-2-2q}{s-q}}\\
&=\sum_{q=0}^\infty\left(\binom{m+1+2q}{q+1}-\binom{m+1+2q}{q}\right)\binom{n-2-2q}{s-q}\tag{1}\\
&=\sum_{q=0}^\infty\left([w^{q+1}]-[w^q]\right)(1+w)^{m+1+2q}[z^{s-q}](1+z)^{n-2-2q}\tag{2}\\
&=\left([w^1]-[w^0]\right)(1+w)^{w+1}[z^s](1+z)^{n-2}\sum_{q=0}^\infty \left(\frac{(1+w)^2}{w}\right)^q\left(\frac{z}{(1+z)^2}\right)^q\tag{3}\\
&=[w^0z^s]\left(\frac{1}{w}-1\right)(1+w)^{m+1}(1+z)^{n-2}\frac{1}{1-\frac{(1+w)^2z}{w(1+z)^2}}\tag{4}\\
&=[w^0z^s](1-w)(1+w)^{m+1}(1+z)^n\frac{1}{w(1+z)^2-(1+w)^2z}\\
&=[w^0z^s](1-w)(1+w)^{m+1}(1+z)^n\frac{1}{w\left(1-\frac{z}{w}\right)(1-wz)}\\
&=[w^0z^s](1-w)(1+w)^{m+1}(1+z)^n\left(\frac{1}{w\left(1-w^2\right)\left(1-\frac{z}{w}\right)}\right.\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left.-\frac{w}{\left(1-w^2\right)(1-wz)}\right)\tag{5}\\
&=[w^0z^s](1+w)^{m}(1+z)^n\left(\frac{1}{w\left(1-\frac{z}{w}\right)}-\frac{w}{1-wz}\right)\tag{6}\\
&=[w^1z^s](1+w)^{m}(1+z)^n\frac{1}{1-\frac{z}{w}}\\
&=[w^1](1+w)^m\sum_{j=0}^s\binom{n}{j}[z^{s-j}]\sum_{k=0}^\infty\left(\frac{z}{w}\right)^k\tag{7}\\
&=[w^1](1+w)^m\sum_{j=0}^s\binom{n}{j}\frac{1}{w^{s-j}}\tag{8}\\
&=[w^{s+1}](1+w)^m\left(\sum_{j=0}^\infty\binom{n}{j}w^j-\sum_{j=s+1}^\infty\binom{n}{j}w^j\right)\\
&=[w^{s+1}](1+w)^m\left((1+w)^n-\sum_{j=s+1}^\infty\binom{n}{j}w^j\right)\\
&\,\,\color{blue}{=\binom{m+n}{s+1}-\binom{n}{s+1}}\tag{9}
\end{align*}
and the claim follows.
The essential steps are (1) where we get rid of the denominator by using a representation as difference of binomial coefficients and the partial fraction decomposition in (5).

Comment:

*

*In (1) we use the binomial identity $\frac{m}{q+1}\binom{m+1+2q}{q}=\binom{m+1+2q}{q+1}-\binom{m+1+2q}{q}$.


*In (2) we apply the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series.


*In (3) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.


*In (4) we use the geometric series expansion and apply again the rule from (3).


*In (5) we do a partial fraction expansion.


*In (6) we observe the right term $\frac{w}{1-wz}=w+w^2z+w^3z^2+\cdots$ can be skipped, since there is no contribution to $[w^0]$.


*In (7) we do again a geometric series expansion and expand the binomial.


*In (8) we select the coefficient of $z^{s-j}$.


*In (9) we select the coefficient of $w^{s+1}$.

OPs first identity:
The right-hand side of the first identity follows from the right-hand side of the second by reversing the order of summation $q\to s-q$.
OPs third identity:
We obtain
\begin{align*}
\color{blue}{\sum_{j=0}^s}&\color{blue}{\frac{m}{m+2j}\binom{m+2j}{j}\binom{n-2j}{s-j}}\\
&=\binom{n}{s}+\sum_{j=1}^s\frac{m}{m+2j}\binom{m+2j}{j}\binom{n-2j}{s-j}\tag{10}\\
&=\binom{n}{s}+\sum_{j=1}^s\frac{m}{j}\binom{m+2j-1}{j-1}\binom{n-2j}{s-j}\tag{11}\\
&=\binom{n}{s}+\sum_{j=0}^{s-1}\frac{m}{j+1}\binom{m+1+2j}{j}\binom{n-2-2j}{s-1-j}\tag{12}\\
&=\binom{n}{s}+\binom{m+n}{s}-\binom{n}{s}\tag{13}\\
&\,\,\color{blue}{=\binom{m+n}{s}}
\end{align*}

Comment:

*

*In (10) we separate the first summand with $j=0$.


*In (11) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.


*In (12) we shift the index and start with $j=0$.


*In (13) we apply (9) by substituting $s$ with $s-1$.
