# 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.

-