How can it be shown that : $$\sum_{r=0}^n\binom {-n-1}r\left(-\frac 12\right)^r=\sum_{r=n+ 1}^\infty\binom {-n-1}r\left(-\frac 12\right)^r=2^n$$ ?
We know that
$$\sum_{r=0}^\infty\binom {-n-1}r\left(-\frac 12\right)^r=\left(1-\frac 12\right)^{-n-1}=2^{n+1}$$ which can be easily confirmed by binomial expansion, but does not seem helpful for solving the above.
Addendum: An algebraic approach would be preferred.
Addendum 2: I arrived at this whilst trying to solve this other question here.
We can use Induction. The base case is simple. Assume it's true for $n-1$, then
$$
\begin{align}
&\sum_{k=0}^n\binom{-n-1}{k}\left(-\frac12\right)^k\\
&=\sum_{k=0}^n\binom{n+k}{k}\left(\frac12\right)^k\tag{1}\\
&=\binom{2n}{n}\left(\frac12\right)^n+\sum_{k=0}^{n-1}\left[\binom{n+k-1}{k-1}+\binom{n+k-1}{k}\right]\left(\frac12\right)^k\tag{2}\\
&=\binom{2n}{n}\left(\frac12\right)^n+\sum_{k=0}^{n-1}\binom{n+k-1}{k-1}\left(\frac12\right)^k+2^{n-1}\tag{3}\\
&=\binom{2n}{n}\left(\frac12\right)^n+\frac12\sum_{k=0}^{n-2}\binom{n+k}{k}\left(\frac12\right)^k+2^{n-1}\tag{4}\\
&=\frac12\sum_{k=0}^n\binom{n+k}{k}\left(\frac12\right)^k+2^{n-1}\tag{5}\\[9pt]
&=2^n\tag{6}
\end{align}
$$
Explanation:
$(1)$: convert to positive binomial coefficient
$(2)$: use Pascal's relation
$(3)$: use inductive hypothesis
$(4)$: reindex
$(5)$: $\binom{2n}{n}\left(\frac12\right)^n=\frac12\binom{2n}{n}\left(\frac12\right)^n+\frac12\binom{2n-1}{n-1}\left(\frac12\right)^{n-1}$
$(6)$: subtract $(1)$ from $2$ times $(5)$
Use the equation ${-n-1\choose r}=(-1)^r{r+n\choose n}$ to rewrite your negative binomial coefficient, and rearrange the desired expression to $$\sum_{r=0}^n{r+n\choose n}\left({1\over 2}\right)^{r+n}=1,$$ or $$\sum_{k=n}^{2n}{k\choose n}\left({1\over 2}\right)^{k}=1.$$
Now consider a probability exercise.
Toss a fair coin until you get either $n+1$ heads or $n+1$ tails, and let $N$ be the number coin tosses needed to achieve this. We get $N=k+1$ by either $n$ heads in the first $k$ tosses followed by a head, or $n$ tails in the first $k$ tosses followed by a tail.
Therefore for $k\leq n\leq 2n$,
$$\mathbb{P}(N=k+1)=2{k\choose n}\left({1\over2}\right)^{k}\left({1\over2}\right)
= {k\choose n}\left({1\over2}\right)^{k}.$$
Since these exhaust all possible values for $N$, the probabilities add up to one.
Here is an answer based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g. \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}
We obtain \begin{align*} \color{blue}{\sum_{r=0}^n}&\color{blue}{\binom{-n-1}{r}\left(-\frac{1}{2}\right)^r}\\ &=\sum_{r=0}^n\binom{n+r}{r}\left(\frac{1}{2}\right)^r\tag{1}\\ &=\sum_{r=0}^n\binom{2n-r}{n-r}\left(\frac{1}{2}\right)^{n-r}\tag{2}\\ &=\sum_{r=0}^\infty[z^{n-r}](1+z)^{2n-r}\left(\frac{1}{2}\right)^{n-r}\tag{3}\\ &=2^{-n}[z^n](1+z)^{2n}\sum_{r=0}^\infty\left(\frac{2z}{1+z}\right)^{r}\tag{4}\\ &=2^{-n}[z^n](1+z)^{2n}\cdot\frac{1}{1-\frac{2z}{1+z}}\tag{5}\\ &=2^{-n}[z^n](1+z)^{2n+1}\cdot\frac{1}{1-z}\tag{6}\\ &=2^{-n}\sum_{k=0}^n[z^k](1+z)^{2n+1}\tag{7}\\ &=2^{-n}\sum_{k=0}^n\binom{2n+1}{k}\tag{8}\\ &=2^{-n}\frac{1}{2}2^{2n+1}\tag{9}\\ &\color{blue}{=2^n} \end{align*}
Comment:
In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q $.
In (2) we change the order of summation $r \rightarrow n-r$.
In (3) we apply the coefficient of operator. We also set the limit to $\infty$ without changing anything since we are adding zeros only.
In (4) we do a rearrangement and apply the formula $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
In (5) we apply the geometric series expansion.
In (6) we do some simplifications.
In (7) we do the Cauchy multiplication with the geometric series $\frac{1}{1-x}$ and restrict the upper limit of the sum with $n$ since other terms do not contribute to $[z^n]$.
In (8) we select the coefficient of $z^k$.
In (9) we apply the binomial theorem.
Here's another approach:
$$\begin{align} \sum_{r=0}^n \binom {-n-1}r\left(-\frac 12\right)^r &=\sum_{r=0}^n\binom {r+n}n2^{-r}\\ &=\frac 1{2^n} \sum_{r=0}^n \binom {r+n}r2^{n-r}\\ &=\frac 1{2^n}\sum_{r=0}^n \binom {r+n}n\sum_{j=0}^{n-r}\binom {n-r}j\\ &=\frac 1{2^n}\sum_{l=0}^n\binom {2n-l}{n-l}\sum_{j=0}^l\binom lj &&(l=n-r)\\ &=\frac 1{2^n}\sum_{j=0}^n\sum_{l=j}^n\binom {2n-l}{n-l}\binom lj\\ &=\frac 1{2^n}\sum_{j=0}^n\binom {2n+1}{n+j+1} &&(*)\\ &=\frac 1{2^n}\cdot \frac 12 \sum_{j=0}^{2n+1}\binom {2n+1}j &&\text{(by symmetry)}\\ &=\frac 1{2^{n+1}}\cdot 2^{2n+1}\\ &=2^n\;\;\color{red}\blacksquare\\\\ ^* \scriptsize \text {using } \sum_r \binom {a+r}c\binom {b+r}d &\scriptsize =\binom {a+b+1}{c+d+1} \end{align}$$
It follows that
$$\begin{align} \sum_{r=n+1}^\infty\binom {-n-1}r\left(-\frac 12\right)^r &=\sum_{r=0}^\infty\binom {-n-1}r\left(-\frac 12\right)^r-\sum_{r=0}^n\binom {-n-1}r\left(-\frac 12\right)^r\\ &=\left(1-\frac 12\right)^{-n-1}-2^n\\ &=2^{n+1}-2^n\\ &=2^n\;\;\color{red}\blacksquare \end{align}$$ Hence $$\sum_{r=0}^n\binom {-n-1}r\left(-\frac 12\right)^r=\sum_{r=n+1}^\infty\binom {-n-1}r\left(-\frac 12\right)^r=2^n$$