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.

This is a nice case where induction loading (the technique of deliberately strengthening what is to be proved for the purpose of making proofs by induction easier) is handy. For motivation see my other answer. Generalise the statement to the claim
$$
  \sum_{k=0}^n\binom{m+n+1}k =\sum_{k=0}^n\binom{m+k}k\times2^{n-k}, \tag1
$$
knowing that for $m=n$ the first sum gives $\sum_{k=0}^{n}\binom{2n+1}k=\frac12\sum_{k=0}^{2n+1}\binom{2n+1}k=2^{2n}$, and one can divide both sides by $2^n$.
Proof of $(1)$ by induction on $n$. For $n=0$ one gets $\binom {m+1}0=1=\binom m0$, so this is fine. So suppose $n>0$ and the result established for $n-1$. We can write $\binom{m+n+1}k=\binom{m+n}k+\binom{m+n}{k-1}$ by Pascal's recurrence (with the convention $\binom a{-1}=0$ for any $a$) so the left hand side is
$$
  \sum_{k=0}^n\left(\binom{m+n}k+\binom{m+n}{k-1}\right)
 =\sum_{k=0}^n\binom{m+n}k+\sum_{l=-1}^{n-1}\binom{m+n}l
\\ =2\sum_{k=0}^{n-1}\binom{m+n}k+\binom{m+n}n.
$$
Now we can apply the induction hypothesis to the summation, and get
$$
  2\sum_{k=0}^{n-1}\binom{m+k}k\times2^{n-1-k}+\binom{m+n}n
=\sum_{k=0}^n\binom{m+k}k\times2^{n-k}
$$
as desired.
