Evaluating the sum $\sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1}$

Question

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 $$\begin{align} \sum_{k=0}^m(-1)^k\binom{n}{k} &=\sum_{k=0}^m(-1)^k\binom{n}{k}\binom{m-k}{m-k}\\ &=(-1)^m\sum_{k=0}^m\binom{n}{k}\binom{-1}{m-k}\\ &=(-1)^m\binom{n-1}{m} \end{align}$$

Answer 1

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.

Answer 2

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

$$[z^n] (1+z)^{2n+1} \sum_{k\ge 0} (-1)^k z^k = [z^n] (1+z)^{2n+1} \frac{1}{1+z} = [z^n] (1+z)^{2n} = {2n\choose n}.$$

Answer 3

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 $\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\frac{\left(1+z\right)^{2n}}{z^{2n+2}}dz=0$.