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.
 A: 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.
A: Note that
$$
\begin{align}
(1+1)^{2n}&=\sum_{k=0}^{2n}\binom{2n}{k}\tag{1}\\
(1-1)^{2n}&=\sum_{k=0}^{2n}\binom{2n}{k}(-1)^k\tag{2}
\end{align}
$$
by the binomial theorem. Subtracting (2) from (1) we get that
$$
2^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}(1-(-1)^k)=2\sum_{0\leq k\leq2n, \, \text{k even}}\binom{2n}{k}=2\sum_{k=0}^{n}\binom{2n}{2k}.
$$
A: The combinatorial proof does not always look out of place. Let us try it here.
The goal : number of subsets of even cardinality , of a set of $2n$ elements. Denote $[m] = \{1,2,...,m\}$.


*

*There is a bijection $\phi$ between even subsets of $[m]$ and any subset of $[m-1]$ : $\phi(A) = A$ if $A$ has an even number of elements, and $\phi(A) = A \cup \{m\}$ otherwise. You can check this is a bijection. Therefore, the number of even subsets of $[m]$ is the number of subsets of $[m-1]$, which is $2^{m-1}$.

*Also, every subset of $[2n]$ has size $2k$ for some $k \leq n$, so the number of such subsets is just the number of possible subsets for some fixed $k$, summed over all possible $k$. This leads to $\sum_{k=0}^n \binom{2n}{2k}$.
Setting $m = 2n$ in the first bullet point and equating the two explanations gives a combinatorial proof of the given identity.
