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.

Let
$$
S_n:=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}
$$
Obviously $S_0=1$. Assume that equality
$$S_{n-1}=1\tag1$$ 
is valid for some $n$. Then it is valid for $n+1$ as well:
$$
\begin{align}
S_n
&=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}\\
&=\sum_{k=0}^n\frac{\binom{n+k-1}{k-1}+\binom{n-1+k}{k}}{2^{n+k}}\\
&=\frac12\sum_{k=0}^{n-1}\frac{\binom{n+k}{k}}{2^{n+k}}+\frac12\sum_{k=0}^n\frac{\binom{n-1+k}{k}}{2^{n-1+k}}\\
&=\frac12S_n-\frac{\binom{2n}{n}}{2^{2n+1}}+\frac12S_{n-1}+\frac{\binom{2n-1}{n}}{2^{2n}}\\
&=\frac12S_n+\frac12S_{n-1}\implies S_n=S_{n-1}\stackrel{I.H.}=1.
\end{align}
$$
Thus, by induction the equality $(1)$ is valid for all integer $n\ge0$.
