Some binomial equality I am trying to prove the following equality
$$
\sum_{r=k}^{n}\binom{2n+1}{2r+1}\binom{r}{k}=\binom{2n-k}{k}2^{2n-2k}~~;~k\le n.
$$
I noticed that for $k=0$ it becomes
$$
\sum_{r=0}^{n}\binom{2n+1}{2r+1}=2^{2n},
$$
which is well known equality. Also have I tried to use the equality 
$$
\sum_{r=k}^{n}\binom{n}{r}\binom{r}{k}=\binom{n}{k}2^{n-k},
$$
but with no success.
Any hints and suggestions would be appreciated.
Thanks!

This  one can  also be  done using  complex variables.  We will  use a
different generating function than in the accepted answer.

Suppose we are trying to evaluate
$$\sum_{r=k}^n {2n+1\choose 2r+1} {r\choose k}.$$
Introduce the integral representation
$${2n+1\choose 2r+1}
= \frac{1}{2\pi i}
\int_{|z|=3/2} \frac{(1+z)^{2n+1}}{z^{2r+2}} \;dz.$$
We will use the annulus $1<|z|<\infty$ with this integral.
This  integral sets  the range  of the  sum so  we can  let $r$  go to
infinity to obtain
$$\frac{1}{2\pi i}
\int_{|z|=3/2} \frac{(1+z)^{2n+1}}{z^2}
\sum_{r=k}^\infty {r\choose k} z^{-2r} \;dz.$$
Observe that the  sum term only converges when  $|z|>1.$ This does not
pose a  problem however as it is contained in the chosen annulus.
The inner sum is
$$\sum_{r=k}^\infty {r\choose k} z^{-2r}
= \sum_{r=0}^\infty {r+k\choose k} z^{-2r-2k}
= z^{-2k} \sum_{r=0}^\infty {r+k\choose k} z^{-2r}
\\ = z^{-2k} \frac{1}{(1-1/z^2)^{k+1}}
= z^2 \times z^{-2(k+1)} \frac{1}{(1-1/z^2)^{k+1}}
\\ = \frac{z^2}{(z^2-1)^{k+1}}.$$
Substitute this into the sum to get
$$\frac{1}{2\pi i}
\int_{|z|=3/2} \frac{(1+z)^{2n+1}}{z^2}
\frac{z^2}{(z^2-1)^{k+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=3/2} \frac{(1+z)^{2n-k}}{(z-1)^{k+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=3/2} \frac{(2+z-1)^{2n-k}}{(z-1)^{k+1}} \; dz
\\ = \frac{2^{2n-k}}{2\pi i}
\int_{|z|=3/2} \frac{(1+(z-1)/2)^{2n-k}}{(z-1)^{k+1}} \; dz.$$
We thus have from the pole at $z=1$
$$2^{2n-k} [(z-1)^k] (1+(z-1)/2)^{2n-k}
\\ = 2^{2n-k} {2n-k\choose k} 2^{-k}
= 2^{2n-2k} {2n-k\choose k}.$$
