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.

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.

