How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$ 
How does one show that $$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$$ for each nonnegative integer $n$?

I tried using the Snake oil technique but I guess I am applying it incorrectly. With the snake oil technique we have $$F(x)= \sum_{n=0}^{\infty}\left\{\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}\right\}x^{n}.$$ I think I have to interchage the summation and do something. But I am not quite comfortable in interchanging the summation. Like after interchaging the summation will $$F(x)=\sum_{k=0}^{n}\sum_{n=0}^{\infty}\binom{n+k}{k}\frac{1}{2^k}x^{n}?$$ Even if I continue with this I am unable to get the correct answer.

*

*How does one prove this using the Snake oil technique?


*A combinatorial proof is also welcome, as are other kinds of proofs.

$$\begin{align}
\sum_{k=0}^n \binom {n+k}k\frac 1{2^k}
&=\frac 1{2^n}\sum_{k=0}^n \color{blue}{2^{n-k}}\binom {n+k}n\\
&=\frac 1{2^n}\sum_{k=0}^n\color{blue}{\sum_{r=0}^{n-k}\binom {n-k}r}\binom{n+k}r\\
&=\frac 1{2^n}\sum_{s=0}^n\sum_{r=0}^s\binom  sr\binom {2n-s}n
&&\scriptsize (\text{Putting  } s=n-k)\\
&=\frac 1{2^n}\sum_{r=0}^n\sum_{s=r}^n \binom s{s-r}\binom{2n-s}{n-s}\\
&=\frac 1{2^n}\sum_{r=0}^n\sum_{s=r}^n(-1)^{s-r}\binom {-r-1}{s-r}(-1)^{n-s}\binom {-n-1}{n-s}
&&\scriptsize(\text{Upper Negation})\\
&=\frac 1{2^n}\sum_{r=0}^n(-1)^{n-r}
\color{magenta}{\sum_{s=r}^n\binom {-r-1}{s-r}\binom{-n-1}{n-s}}\\
&=\frac 1{2^n}\sum_{r=0}^n(-1)^{n-r}\color{magenta}{\binom {-n-r-2}{n-r}}&&\scriptsize(\text{Vandermonde})\\
&=\frac 1{2^n}\sum_{r=0}^n(-1)^{n-r}(-1)^{n-r}\binom {2n+1}{n-r}&&\scriptsize(\text{Upper Negation})\\
&=\frac 1{2^n}\sum_{r=0}^n\binom{2n+1}{n-r}\\
&=\frac 1{2^n}\cdot \frac 12 \sum_{r=0}^n \binom {2n+1}{n-r}+\binom {2n+1}{n+r+1}
&&\scriptsize(\text{both summands are equal})\\
&=\frac 1{2^{n+1}}\sum_{r=0}^{2n+1}\binom {2n+1}r\\
&=\frac 1{2^{n+1}}\cdot 2^{2n+1}\\\\
&=2^n\qquad \blacksquare
\end{align}$$
