Partial Sum of Negative Binomial Coefficients multiplied by Rising Power of $-\frac 12$ 
How can it be shown that :
  $$\sum_{r=0}^n\binom {-n-1}r\left(-\frac 12\right)^r=\sum_{r=n+
1}^\infty\binom {-n-1}r\left(-\frac 12\right)^r=2^n$$
  ?
This is confirmed by wolframalpha here and here.

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. 
 A: 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 $(5)$
A: 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.
A: 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.
A: 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 
$$\sum_{r=0}^n\binom {-n-1}r\left(-\frac 12\right)^r=\sum_{r=n+1}^\infty\binom {-n-1}r\left(-\frac 12\right)^r=2^n$$
