Proof of $\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}$ I found some combinatorial identities in my old notebooks, but I cannot recall how I derived them. Can anyone help provide the most elementary/elegant proofs for the following identity? Specifically, the connection between the summation and either of the two closed-form expressions.
$$
\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}
$$
Some specific cases written out for some simple visualization: ($n=0$ through $n=3$)
$$
\begin{align}
\binom{0}{0} 2^{0} &= 1\binom{0}{0} = \frac{1}{2}\binom{2}{1} \\
\binom{0}{0} 2^{2} + \binom{2}{1} 2^{0} &= 3\binom{2}{1}  = \frac{2}{2}\binom{4}{2} \\
\binom{0}{0} 2^{4} + \binom{2}{1} 2^{2} + \binom{4}{2} 2^{0} &= 5\binom{4}{2} = \frac{3}{2}\binom{6}{3}\\
\binom{0}{0} 2^{6} + \binom{2}{1} 2^{4} + \binom{4}{2} 2^{2} + \binom{6}{3} 2^{0} &= 7\binom{6}{3} = \frac{4}{2}\binom{8}{4}
\end{align}
$$
EDIT: I should specify that I have already found an inductive proof, but it is not satisfying to me. I am specifically looking for a bijective/double-counting proof ideally.

Proving the equality between the two closed form is easier:
$$\begin{align*}
(2n+1)\binom{2n}n &=(2n+1)\frac{(2n)!}{n!n!}\\
&= \frac{(2n+1)!}{n!n!}\\
&= \frac{(n+1)^2}{(2n+2)}\cdot\frac{(2n+2)!}{(n+1)!(n+1)!}\\
&= \frac{n+1}{2}\binom{2(n+1)}{n+1}
\end{align*}$$
By induction, consider the $n+1$ case of the summation form and the second closed-form,
$$\begin{align*}
\sum_{k=0}^{n+1} \binom{2k}{k}2^{2(n+1)-2k} &= 2^2\sum_{k=0}^{n}\binom{2k}{k}2^{2n-2k} + \binom{2(n+1)}{n+1}2^0\\
&= 2^2\cdot \underbrace{\frac{n+1}{2} \binom{2(n+1)}{n+1)}}_{\text{by induction hypothesis}} + \binom{2(n+1)}{n+1}2^0\\
&= [2(n+1)+1]\binom{2(n+1)}{n+1}
\end{align*}$$
