$\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$: 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?

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_{k=0}^n\binom{n+k}{k}\frac{1}{2^{n+k}}}
&=\sum_{k=0}^n\binom{2n-k}{n-k}\frac{1}{2^{2n-k}}\tag{1}\\
&=\sum_{k=0}^\infty[z^{n-k}](1+z)^{2n-k}\frac{1}{2^{2n-k}}\tag{2}\\
&=2^{-2n}[z^n](1+z)^{2n}\sum_{k=0}^\infty\left(\frac{2z}{1+z}\right)^{k}\tag{3}\\
&=2^{-2n}[z^n](1+z)^{2n}\cdot\frac{1}{1-\frac{2z}{1+z}}\tag{4}\\
&=2^{-2n}[z^n](1+z)^{2n+1}\cdot\frac{1}{1-z}\tag{5}\\
&=2^{-2n}\sum_{k=0}^n[z^k](1+z)^{2n+1}\tag{6}\\
&=2^{-2n}\sum_{k=0}^n\binom{2n+1}{k}\tag{7}\\
&=2^{-2n}\frac{1}{2}2^{2n+1}\tag{8}\\
&\color{blue}{=1}
\end{align*}

Comment:


*

*In (1) we change the order of summation $k \rightarrow n-k$.

*In (2) we apply the coefficient of operator. We also set the limit to $\infty$ without changing anything since we are adding zeros only.

*In (3) we do a rearrangement and apply the formula $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (4) we apply the geometric series expansion.

*In (5) we do some simplifications.

*In (6) 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 (7) we select the coefficient of $z^k$.

*In (8) we apply the binomial theorem.
