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.
Returning to the start we seek
$$\sum_{q=0}^K (-1)^q {2n+1-q\choose q} {2n-2q\choose K-q} = \frac{1}{2} (1+(-1)^K).$$
With
$${2n-2q\choose K-q} = {2n-2q\choose 2n-K-q} = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{K-q+1}} \frac{1}{(1-w)^{2n-K-q+1}} \; dw$$
the LHS becomes
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{2\pi i} \int_{|z|=\varepsilon} \sum_{q=0}^K (-1)^q \frac{1}{z^{q+1}} (1+z)^{2n+1-q} \frac{1}{w^{K-q+1}} \frac{1}{(1-w)^{2n-K-q+1}} \; dz \; dw.$$
Here we may extend the sum to infinity due to the pole at $w=0$ vanishing when $q\gt K.$ This yields
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{K+1}} \frac{1}{(1-w)^{2n-K+1}} \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z} (1+z)^{2n+1} \\ \times \sum_{q\ge 0} (-1)^q \frac{1}{z^q} (1+z)^{-q} w^q (1-w)^{q} \; dz \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{K+1}} \frac{1}{(1-w)^{2n-K+1}} \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z} (1+z)^{2n+1} \\ \times \frac{1}{1+w(1-w)/z/(1+z)} \; dz \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{K+1}} \frac{1}{(1-w)^{2n-K+1}} \frac{1}{2\pi i} \int_{|z|=\varepsilon} (1+z)^{2n+2} \\ \times \frac{1}{z(1+z)+w(1-w)} \; dz \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{K+1}} \frac{1}{(1-w)^{2n-K+1}} \frac{1}{2\pi i} \int_{|z|=\varepsilon} (1+z)^{2n+2} \\ \times \frac{1}{(z+w)(z-(w-1))} \; dz \; dw.$$
Once more the pole at $z=0$ is gone but a new one has appeared (two of them, in fact). To see this note that in the summation of the infinite series we require for convergence that $|w(1-w)| \lt |z(1+z)|.$ We have for $\varepsilon \ll 1$ and $\gamma \ll 1$ that $|z(1+z)|\ge \varepsilon (1-\varepsilon) \ge \frac{1}{2} \varepsilon$ and $|w(1-w)| \le \gamma (1+\gamma) \le \frac{3}{2}\gamma.$ Hence $3\gamma \lt \varepsilon$ will work e.g. take $\gamma = \varepsilon/4$ as in $\gamma = 1/16$ and $\varepsilon = 1/4.$ In particular $|w|\lt |z|$ so the pole at $z_0=-w$ is inside the contour. On the other hand the closest that the pole at $z_1=w-1$ which is on a circle at $z=-1$, rotating with radius $\gamma$, gets to the origin is $1-\gamma = 1 - \varepsilon/4 \gt \varepsilon$ as long as $\varepsilon \lt 4/5$, so this is definitely not inside the contour. Hence the contribution from $z_0 = -w$ is the only one and it yields
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{K+1}} \frac{1}{(1-w)^{2n-K+1}} (1-w)^{2n+2} \frac{1}{1-2w} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{K+1}} (1-w)^{K+1} \frac{1}{1-2w} \; dw \\ = \sum_{q=0}^K {K+1\choose q} (-1)^q 2^{K-q} = - (-1)^{K+1} \frac{1}{2} + \frac{1}{2} \sum_{q=0}^{K+1} {K+1\choose q} (-1)^q 2^{K+1-q} \\ = - (-1)^{K+1} \frac{1}{2} + \frac{1}{2} 1^{K+1} = \frac{1}{2} (1 + (-1)^K).$$
We once more have the claim.
Here is a starter. We transform the sum and separate one part which can be simplified. We start with the left hand side of OPs identity and obtain \begin{align*} \sum_{i=0}^K&(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}\\ &=\sum_{i=0}^K(-1)^i\frac{(2n+1-i)!}{i!(2n+1-2i)!}\,\frac{(2n-2i)!}{(K-i)!(2n-K-i)!}\\ &=\sum_{i=0}^K(-1)^i\binom{K}{i}\binom{2n-i}{K}\frac{2n+1-i}{2n+1-2i}\tag{1}\\ \end{align*}
With (1) we can reformulate OPs claim for $0\leq K\leq 2n$ as \begin{align*} \color{blue}{\sum_{i=0}^K(-1)^i\binom{K}{i}\binom{2n-i}{K}\frac{2n+1-i}{2n+1-2i}=\frac{1}{2}\left(1+(-1)^K\right)}\tag{2} \end{align*}
We can write the left-hand side of (2) as \begin{align*} \sum_{i=0}^K(-1)^i\binom{K}{i}\binom{2n-i}{K}\left(1+\frac{i}{2n+1-2i}\right)\tag{3} \end{align*} In the following we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance \begin{align*} [z^p](1+z)^q=\binom{q}{p}\tag{4} \end{align*}
We obtain from (3) \begin{align*} \color{blue}{\sum_{i=0}^K}&\color{blue}{(-1)^i\binom{K}{i}\binom{2n-i}{K}}\\ &=\sum_{i=0}^K(-1)^i\binom{K}{k}\binom{2n-i}{2n-i-K}\tag{5.1}\\ &=\sum_{i=0}^K(-1)^i\binom{K}{i}[z^{2n-i-K}](1+z)^{2n-i}\tag{5.2}\\ &=[z^{2n-K}](1+z)^{2n}\sum_{i=0}^K\binom{K}{i}\left(-\frac{z}{1+z}\right)^i\tag{5.3}\\ &=[z^{2n-k}](1+z)^{2n}\left(1-\frac{z}{1+z}\right)^K\tag{5.4}\\ &=[z^{2n-k}](1+z)^{2n-K}\\ &=\binom{2n-K}{2n-K}\\ &\,\,\color{blue}{=1}\tag{5.5} \end{align*}
Comment:
In (5.1) we use $\binom{p}{q}=\binom{p}{p-q}$.
In (5.2) we use (3).
In (5.3) we use $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
In (5.4) we apply the binomial theorem and do some simplifications in the following steps.
We also obtain from (3) \begin{align*} \color{blue}{\sum_{i=1}^K}&\color{blue}{(-1)^i\binom{K}{i}\binom{2n-i}{K}\frac{i}{2n+1-2i}}\\ &=K\sum_{i=1}^K(-1)^i\binom{K-1}{i-1}\binom{2n-i}{K}\frac{1}{2n+1-2i}\tag{6.1}\\ &=K\sum_{i=0}^{K-1}(-1)^{i+1}\binom{K-1}{i}\binom{2n-i-1}{K}\frac{1}{2n-1-2i}\tag{6.2}\\ &=K\sum_{i=0}^{K-1}(-1)^{i+1}\binom{K-1}{i}\binom{2n-i-1}{2n-i-1-K}\frac{1}{2n-1-2i}\tag{6.3}\\ &\color{blue}{=(-1)^KK\sum_{i=0}^{K-1}\binom{K-1}{i}\binom{-K-1}{2n-i-1-K}\frac{1}{2n-1-2i}}\tag{6.4}\\ \end{align*}
Comment:
In (6.1) we use $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
In (6.2) we shift the index to start with $i=0$.
In (6.3) we use $\binom{p}{q}=\binom{p}{p-q}$.
In (6.4) we use $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
We finally derive from (3), (5.5) and (6.4) the following claim \begin{align*} \color{blue}{K\sum_{i=0}^{K-1}\binom{K-1}{i}\binom{-K-1}{2n-i-1-K}\frac{1}{2n-1-2i}=\frac{1}{2}\left(1-(-1)^K\right)} \end{align*} which could alternatively be used to prove OPs claim.
