Prove $2 \cdot \sum\limits_{k = 0}^{n} \binom{2n}{2k} = 2^{2n}$ without induction I  am aware of
$$
2^{2n} = (1 + 1)^{2n} =
\sum_{k = 0}^{2n} \binom{2n}{k}
$$
I then tried grouping the terms and using that $\binom{n}{k} = \binom{n}{n-k}$, to obtain
$$
\begin{align}
\sum_{k = 0}^{2n} \binom{2n}{k} &=
\binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{n - 1} + \binom{2n}{n} + \binom{2n}{n + 1} + \ldots + \binom{2n}{2n} \\
&= 
2 \left[ \binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{n - 1}\right] + \binom{2n}{n} =
\left( 2 \sum_{k = 0}^{n-1} \binom{2n}{k} \right) + \binom{2n}{n}
\end{align}
$$
But I don't know if I'm on the right track or how to progress from here.

I would also use
$$0 = (1 - 1)^{2n} = \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}.$$
Adding this to the previous sum,
$$2^{2n} = \sum_{k=0}^{2n}\left[(-1)^k\binom{2n}{k} + \binom{2n}{k}\right].$$
When $k$ is odd, the term is $0$, leaving only even $k$ terms (but doubled!). Therefore,
$$2 \sum_{k=0}^{n} \binom{2n}{2k} = 2^{2n},$$
as required.
