# Proving combinatorial identity $\sum_s (-1)^s\binom{p+s-1}{s}\binom{2m+2p+s}{2m+1-s}2^s=0$

I need to prove following combinatorial identities:

$$\sum\limits_s(-1)^s\binom{p+s-1}{s}\binom{2m+2p+s}{2m+1-s}2^s=0$$

$$\sum\limits_s(-1)^s\binom{p+s-1}{s}\binom{2m+2p+s-1}{2m-s}2^s=(-1)^m\binom{p+m-1}{m}$$

given the fact that

$$(1-x)^{2k}\left(1+\frac{2x}{(1-x)^2}\right)^k = (1+x^2)^k$$

for either $k=p$ or $k=-p$.

And for the first one I cannot understand where $\binom{2m+2p+s}{2m+1-s}$ is emerging from (different signs for $s$ on top and bottom seem strange to me). I'm trying to prove first identity as following: if we transform given equation and let $k=p$ we get something like this:

$$(1-x)^{2p}(1+2x+4x^2+\dots)^p=(1+x^2)^p$$

Let's find coefficient for $x^{2m+1}$ for both sides. For the right side it is always equal to $0$, as only even powers are present there. For the left side let's take $x^s$ from second bracket and $x^{2m+1-s}$ from first. Getting $x^s$ from second bracket is equal to splitting $s$ into $p$ addends with zeroes allowed, so the coefficient is equal to $2^s\binom{p+s-1}{p-1} = 2^s\binom{p+s-1}{s}$. So we get some of needed multipliers for our identity. But now if we take $2m+1-s$ from first bracket we get coefficient like $(-1)^{2m+1-s}\binom{2p}{2m+1-s} = (-1)^s\binom{2p}{2m+1-s}$. And the final result is:

$$\sum\limits_s(-1)^s\binom{p+s-1}{s}\binom{2p}{2m+1-s}2^s=0$$

And I see no way to transform it to required identity.

For the second equality I do not even understand where right side is taken from.

Thanks in advance for any help.

Ok, first of all there was an error in my attempt to solve, because of my misinterpreting of series $1+2x+4x^4+\dots$ as $\sum(2x)^n$ when actually (if one to find more series members) it is $1 + 2x + 4x^2 + 6x^3 + 8x^4+\dots$. Secondly, this attempt was incorrect way of solving this problem. The correct way involves applying binomial theorem, like this:

$$(1-x)^{2k}\left(1+\dfrac{2x}{(1-x)^2}\right)^k = (1-x)^{2k} \sum\limits_s \binom{k}{s} \left(\dfrac{2x}{(1-x)^2}\right)^s = \sum\limits_s\binom{k}{s}2^sx^s(1-x)^{2k-2s}$$

After that we can apply it second time:

$$\sum\limits_s\binom{k}{s}2^sx^s(1-x)^{2k-2s} = \sum\limits_s\binom{k}{s}2^sx^s\sum\limits_t(-1)^t\binom{2k-2s}{t}x^t = \sum\limits_s\sum\limits_t2^s(-1)^t\binom{k}{s}\binom{2k-2s}{t}x^{s+t}$$

And now one should just take $k=-p$ and $s+t=2m$ for even powers and $s+t=2m+1$ for odd ones. This directly leads to desired identities.

Suppose we seek to verify that $$\sum_{q=0}^{2m} (-1)^q {p-1+q\choose q} {2m+2p+q-1\choose 2m-q} 2^q = (-1)^m {p-1+m\choose m}.$$

Introduce $${2m+2p+q-1\choose 2m-q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2m-q+1}} (1+z)^{2m+2p+q-1} \; dz.$$

Observe that this controls the range being zero when $q\gt 2m$ so we may extend $q$ to infinity to obtain for the sum

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2m+1}} (1+z)^{2m+2p-1} \sum_{q\ge 0} {p-1+q\choose q} (-1)^q 2^q z^q (1+z)^q \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2m+1}} (1+z)^{2m+2p-1} \frac{1}{(1+2z(z+1))^p} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2m+1}} (1+z)^{2m+2p-1} \frac{1}{((1+z)^2+z^2)^p} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2m+1}} (1+z)^{2m-1} \frac{1}{(1+z^2/(1+z)^2)^p} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2m}} (1+z)^{2m} \frac{1}{z(1+z)} \frac{1}{(1+z^2/(1+z)^2)^p} \; dz.$$

Now put $$\frac{z}{1+z} = u \quad\text{so that}\quad z=\frac{u}{1-u} \quad\text{and}\quad dz=\frac{1}{(1-u)^2} du$$

to obtain for the integral $$\frac{1}{2\pi i} \int_{|u|=\epsilon} \frac{1}{u^{2m}} \frac{1}{u/(1-u)\times 1/(1-u)} \frac{1}{(1+u^2)^p} \frac{1}{(1-u)^2} \; du \\ = \frac{1}{2\pi i} \int_{|u|=\epsilon} \frac{1}{u^{2m+1}} \frac{1}{(1+u^2)^p} \; du.$$

This is $$[u^{2m}] \frac{1}{(1+u^2)^p} = [v^{m}] \frac{1}{(1+v)^p} = (-1)^m {m+p-1\choose m},$$

as claimed.