$\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$: short proof? The identity $\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$ arises from a question on probability in my textbook. A proof by induction on $n$, which exploits the fact that $\binom{a}{b}+\binom{a}{b+1}=\binom{a+1}{b+1}$, is straightforward but not enlightening.
Is it possible to find any very clever approaches? Via a combinatorial or probabilistic interpretation, for instance?

Updated Solution
Here's a neater solution!
$$\begin{align}
\sum_{k=0}^n \frac 1{2^{n+k}}\binom {n+k}n
&=\frac 1{2^{2n}}\sum_{k=0}^n \binom {k+n}k 2^{n-k}\\
&=\frac 1{2^{2n}}\sum_{k=0}^n \binom {k+n}k \sum_{j=0}^{n-k}\binom {n-k}j\\
&=\frac 1{2^{2n}}\sum_{l=0}^n \binom {2n-l}{n-l}\sum_{j=0}^l \binom lj
&&(l=n-k)\\
&=\frac 1{2^{2n}}\sum_{j=0}^n \sum_{l=j}^n\binom {2n-l}n\binom lj\\
&=\frac 1{2^{2n}}\sum_{j=0}^n \binom {2n+1}{n+j+1}
&&(*)\\
&=\frac 1{2^{2n}}\sum_{j=n+1}^{2n+1} \binom {2n+1}j\\
&=\frac 1{2^{2n}}\cdot \frac 12\sum_{j=0}^{2n+1}\binom {2n+1}j
&&\text{(by symmetry)}\\
&=\frac 1{2^{2n+1}}\cdot 2^{2n+1}\\
&=1\;\;\color{red}{\blacksquare}\end{align}$$
$\qquad \qquad \quad ^*\displaystyle\scriptsize\text{using }\sum_{r} \binom {a-r}{c}\binom {b+r}{d}=\binom {a+b+1}{c+d+1}$

Solution posted earlier 
Here's a direct algebraic proof without using induction. 
$$\begin{align}
\sum_{k=0}^n \binom {k+n}k x^k(x+y)^{n-k}
&=\sum_{k=0}^n \binom {k+n}k x^k\sum_{j=0}^{n-k}\binom {n-k}Jy^jx^{n-k-j}\\
&=\sum_{k=0}^n \binom{k+n}k\sum_{j=0}^{n-k}\binom{n-k}jy^jx^{n-j}\\
&=\sum_{\ell=0}^n\binom{2n-\ell}{n-\ell}\sum_{j=0}^\ell\binom{\ell}jy^jx^{n-j}
&&(\ell=n-k)\\
&=\sum_{j=0}^n\sum_{\ell=j}^n (-1)^{n-\ell}\binom{-n-1}{n-\ell}(-1)^{\ell-j}\binom{-j-1}{\ell-j}y^jx^{n-j}\\
&=\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\sum_{\ell=j}^n\binom{-n-1}{n-\ell}\binom{-j-1}{\ell-j}\\
&=\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\binom{-n-j-2}{n-j}
&&\text{(Vandermonde)}\\
&=\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\cdot (-1)^{n-j}\binom{2n+1}{n-j}\\
&=\sum_{j=0}^n \binom{2n+1}{n-j}y^jx^{n-j}\\
&=\sum_{i=0}^n \binom{2n+1}i x^i y^{n-i}\\
\text{Put }x=y=1:\hspace{4cm}\\
\sum_{k=0}^n\binom {k+n}k2^{n-k}
&=\sum_{i=0}^m\binom{2n+1}i
&&(i=n-j)\\
&=\frac 12\cdot 2^{2n+1}
&&\text{(by symmetry)}\\
&=2^{2n}\\
\sum_{k=0}^n \binom {n+k}k 2^{-k}&=2^n\\
\sum_{k=0}^n\frac 1{2^{n+k}} \binom {n+k}n&=1\\
\end{align}$$
