Evaluating the sum $\sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1}$ I encountered this sum in Sasha's answer to this question and I have no idea how to approach it. I don't understand  or even see the telescoping trick mentioned in the answer. I also don't understand the last two lines of this $$\begin{align}
\sum_{k=0}^m(-1)^k\binom{n}{k}
&=\sum_{k=0}^m(-1)^k\binom{n}{k}\binom{m-k}{m-k}\\
&=(-1)^m\sum_{k=0}^m\binom{n}{k}\binom{-1}{m-k}\\
&=(-1)^m\binom{n-1}{m}
\end{align}$$
 A: 
Telescoping sum: We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{2n+1}{n+k+1}}\\
&=\sum_{k=0}^n(-1)^k\frac{2n+1}{n+k+1}\binom{2n}{n+k}\tag{1}\\
&=\sum_{k=0}^n(-1)^k\left(\frac{n-k}{n+k+1}+1\right)\binom{2n}{n+k}\\
&=\sum_{k=0}^n(-1)^k\frac{n-k}{n+k+1}\binom{2n}{n+k}+\sum_{k=0}^n(-1)^k\binom{2n}{n+k}\\
&=\sum_{k=0}^n(-1)^k\binom{2n}{n+k+1}-\sum_{k=0}^n(-1)^{k-1}\binom{2n}{n+k}\tag{2}\\
&=\sum_{k=0}^n(-1)^k\binom{2n}{n+k+1}-\sum_{k=-1}^{n-1}(-1)^{k}\binom{2n}{n+k+1}\tag{3}\\
&=(-1)^n\binom{2n}{2n+1}-(-1)^{-1}\binom{2n}{n}\tag{4}\\
&\,\,\color{blue}{=\binom{2n}{n}}
\end{align*}

Comment:

*

*In (1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.


*In (2) we use the binomial identity $\binom{p}{q+1}=\frac{p!}{(q+1)!(p-q-1)!}=\frac{p-q}{q+1}\frac{p!}{q!(p-q)!}=\frac{p-q}{q+1}\binom{p}{q}$.


*In (3) we  shift the index of the right-hand sum  by $1$ to start with $k=-1$.


*In (4) we have $g(n)-g(-1)$ with $g(k)=(-1)^k\binom{2n}{n+k+1}$.
Note $\binom{p}{q}=0$ if  $q>p$.

We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^m(-1)^k\binom{n}{k}}
&=\sum_{k=0}^m(-1)^k\binom{n}{k}\binom{m-k}{m-k}\tag{5}\\
&=(-1)^m\sum_{k=0}^m\binom{n}{k}\binom{-1}{m-k}\tag{6}\\
&\,\,\color{blue}{=(-1)^m\binom{n-1}{m}}\tag{7}
\end{align*}

Comment:

*

*In (5) we  multiply by $1=\binom{m-k}{m-k}$.


*In (6) use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.


*In (7) we apply the Chu-Vandermonde Identity.
A: We seek to evaluate
$$\sum_{k=0}^n (-1)^k {2n+1\choose n+k+1}
= \sum_{k=0}^n (-1)^k {2n+1\choose n-k}
\\ = \sum_{k=0}^n (-1)^k [z^{n-k}] (1+z)^{2n+1}
= [z^n]  (1+z)^{2n+1} \sum_{k=0}^n (-1)^k z^k.$$
Now the coefficient extractor $[z^n]$ combined with $z^k$ enforces the
range (no contribution when $k\gt n$) and we may continue with
$$[z^n]  (1+z)^{2n+1} \sum_{k\ge 0} (-1)^k z^k
= [z^n]  (1+z)^{2n+1} \frac{1}{1+z}
= [z^n]  (1+z)^{2n} = {2n\choose n}.$$
A: From the integral representation of the binomial coefficient $$\dbinom{n}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\frac{\left(1+z\right)^{n}}{z^{k+1}}dz$$ we have $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{2n+1}{n+k+1}=\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\left(1+z\right)^{2n+1}\sum_{k=0}^{n}\left(-1\right)^{k}\frac{1}{z^{n+k+2}}dz$$ $$=\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\left(1+z\right)^{2n}\left(z^{-n-1}+\left(-1\right)^{n}z^{-2n-2}\right)dz$$ $$=\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\frac{\left(1+z\right)^{2n}}{z^{n+1}}dz+\left(-1\right)^{n}\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\frac{\left(1+z\right)^{2n}}{z^{2n+2}}dz=\dbinom{2n}{n}$$ since $\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\frac{\left(1+z\right)^{2n}}{z^{2n+2}}dz=0$.
