Prove ${2n\choose0} + {2n\choose2} + \dots + {2n\choose{2n}} = 2^{2n-1}$ 
Prove the identity:
  $$
{2n\choose0} + {2n\choose2} + \dots + {2n\choose{2n}} = 2^{2n-1}
$$  
$\forall n\in\mathbb{N}$  

Progress:  
I've tried using the binomial theorem to get:
$$
2^{2n-1}=\sum\limits_{k=0}^{2n-1} {{2n-1}\choose k}
$$
Followed by Pascal's:
$$
{2n\choose k}={{2n-1}\choose k}+{{2n-1}\choose{k-1}}
$$ Then:
$$
2^{2n-1}=\sum\limits_{k=0}^{2n-1} \Bigg({2n\choose k}-{{2n-1}\choose{k-1}}\Bigg)
$$ However, I don't feel right about this.
 A: Consider $2^{2n} = (1+1)^{2n} = {2n\choose0} + {2n\choose1} + {2n\choose2} \dots + {2n\choose{2n}}$.
$0 = (1+(-1))^{2n} = {2n\choose0} - {2n\choose1} + {2n\choose2} \dots - {2n\choose{2n-1}} + {2n\choose{2n}}$. Now add these two equations and we get 
$2^{2n} = 2({2n\choose0} + {2n\choose2} + \dots + {2n\choose{2n}})$. Therefore, ${2n\choose0} + {2n\choose2} + \dots + {2n\choose{2n}} = 2^{2n-1}$.
A: The left side is the number of subsets in $\{1,2,...,2n\}$ with even power and the number of them is half of all subsets that is right side.
A: Here's a combinatorial proof. Let $E$ be the set of subsets of $\{1,2,\dots,2n\}$ containing an even number of elements. Then


*

*$E$ can be partitioned as $E_0 \cup E_2 \cup \cdots \cup E_{2n}$, where $E_{k}$ is the set of subsets of size $k$.
$$|E| = \binom{2n}{0} + \binom{2n}{2} + \cdots + \binom{2n}{2n}$$

*$E$ is in bijection with the set $O$ of subsets of odd size (proving this is a good exercise: the idea behind defining the bijection $E \to O$ is to remove $2n$ from a subset if it is an element and to add $2n$ if it is not). But then we have $|E|=|O|$, and so
$$2^{2n} = |\mathcal{P}(\{1,2,\dots,2n\})| = |E \cup O| = |E|+|O| = 2|E|$$
so that $|E|=2^{2n-1}$


Putting this together yields
$$\binom{2n}{0} + \binom{2n}{2} + \cdots + \binom{2n}{2n} = 2^{2n-1}$$
as required.
A: You know that
$$
(1+x)^n=\sum_{i=0}^n \binom{n}{i}x^i.
$$
Fix $x=1$ and $x=-1$ and you get
$$
2^n=\sum_{i=0}^n \binom{n}{i} \,\,\,\text{ and }\,\,\,\sum_{\substack{0\le i\le n}\\i\text{ even } }\binom{n}{i}=\sum_{\substack{0\le i\le n}\\i\text{ odd } } \binom{n}{i}.
$$
Therefore
$$
\sum_{\substack{0\le i\le n}\\i\text{ even } }\binom{n}{i}=\sum_{\substack{0\le i\le n}\\i\text{ odd } } \binom{n}{i}=2^{n-1}.
$$
