Proving $\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$ I encountered the following binomial equality:
$$\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$$
which I know it's true, but I don't know how to prove it directly. I entered the left-hand-side to Mathematica, and it directly gave me the right-hand-side. So I wonder if anyone knows an elementary proof.

We seek to show that
$$\sum_{q=0}^K (-1)^q {2n+1-q\choose q}
{2n-2q\choose K-q} = \frac{1}{2} (1+(-1)^K).$$
The LHS is
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\sum_{q=0}^K
(-1)^q \frac{1}{z^{q+1}} (1+z)^{2n+1-q}
\frac{1}{w^{K-q+1}} (1+w)^{2n-2q}
\; dz \; dw
\\ = \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{1}{z} (1+z)^{2n+1}
\\ \times
\sum_{q=0}^K
(-1)^q \frac{1}{z^q} (1+z)^{-q}
w^q (1+w)^{-2q}
\; dz \; dw.$$
Here we may extend $q$ beyond $K$ to infinity because the pole at zero
in $w$ is canceled for the extra values. We obtain
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{1}{z} (1+z)^{2n+1}
\\ \times
\frac{1}{1+w/(1+w)^2/z/(1+z)}
\; dz \; dw
\\ = \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n+2}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
(1+z)^{2n+2}
\\ \times
\frac{1}{(1+z(1+w))(w+z(1+w))}
\; dz \; dw.$$
The pole at zero in $z$ is gone but a new pole has appeared inside the
contour. Note that when we summed the geometric series we required
$|w/(1+w)^2| \lt |z(1+z)|.$ We have with $\gamma\ll 1$ and
$\varepsilon\ll 1$ that $|w/(1+w)^2| \le  \gamma/(1-\gamma)^2 \lt
2\gamma$ and $|z(1+z)| \ge \varepsilon (1-\varepsilon) \gt
\frac{1}{2}\varepsilon.$ Therefore  taking $\varepsilon = 4\gamma$ will
work e.g. $\gamma=1/11$ and  $\varepsilon = 4/11.$
We have for the first simple pole at  $z_0=-1/(1+w)$ that $|-1/(1+w)| \gt
1/(1+\gamma) \gt 4\gamma =  \varepsilon.$ This pole is not inside the
contour. The second pole is  at $z_1=-w/(1+w)$ and we have $|-w/(1+w)|
\lt \gamma/(1-\gamma) \lt  4\gamma = \varepsilon.$ This pole is inside
the contour. We thus write
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n+1}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
(1+z)^{2n+2}
\\ \times
\frac{1}{(1+z(1+w))(w/(1+w)+z)}
\; dz \; dw.$$
Evaluating the residue from the simple pole at $z_1$ we find
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n+1}
(1-w/(1+w))^{2n+2} \frac{1}{1-(1+w)w/(1+w)} \; dw
\\ = \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} \frac{1}{1-w^2} \; dw.$$
This is
$$[w^K] \frac{1}{1-w^2} = \frac{1}{2}(1+(-1)^K)$$
as claimed.
