How prove this $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1 $ Show that
$$\sum_{k=0}^{n}\dfrac{\binom{2n-k}{n}}{2^{2n-k}}=1$$
I think this problem can be solved with nice methods, such as algebraic ones. Or can I use probability methods? Thank you 

Suppose we seek to show that
$$\sum_{k=0}^{n} {2n-k\choose n} 2^k = 2^{2n}.$$
Introduce
$${2n-k\choose n}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-k}}{z^{n+1}} \; dz.$$
This gives for the sum that
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} 
\sum_{k=0}^n \frac{2^k}{(1+z)^k}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} 
\frac{1-2^{n+1}/(1+z)^{n+1}}{1-2/(1+z)} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} 
\frac{1+z-2^{n+1}/(1+z)^{n}}{1+z-2} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} 
\frac{1+z-2^{n+1}/(1+z)^{n}}{z-1} \; dz.$$
The difference here yields two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} 
\frac{2^{n+1}/(1+z)^{n}}{1-z} \; dz
\\ = \frac{2^{n+1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}} 
\frac{1}{1-z} \; dz.$$
This yields
$$2^{n+1} \sum_{q=0}^n {n\choose q} = 2^{2n+1}.$$
The second piece is
$$- \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} 
\frac{1+z}{1-z} \; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}} 
\frac{1}{1-z} \; dz.$$
This is
$$-\sum_{q=0}^{n} {2n+1\choose q}
= -\frac{1}{2} 2^{2n+1} = - 2^{2n}.$$
Combining the two pieces we have
$$2^{2n+1} - 2^{2n} = 2^{2n}$$
as claimed.
