Prove ${2n\choose0} + {2n\choose2} + \dots + {2n\choose{2n}} = 2^{2n-1}$

Question

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.

See also: Evaluate $ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots$ and other posts linked there. — Martin Sleziak, Dec 04 '17 at 18:56

score 2 · Accepted Answer · answered Dec 04 '17 at 17:16

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, ${2n\choose0} + {2n\choose2} + \dots + {2n\choose{2n}} = 2^{2n-1}$.

score 0 · Answer 2 · answered Dec 04 '17 at 17:04

0

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.

answered Dec 04 '17 at 17:04

nonuser

90,026

score 0 · Answer 3 · answered Dec 04 '17 at 17:07

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.

score 0 · Answer 4 · answered Dec 04 '17 at 17:13

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 $$ \sum_{\substack{0\le i\le n}\\i\text{ even } }\binom{n}{i}=\sum_{\substack{0\le i\le n}\\i\text{ odd } } \binom{n}{i}=2^{n-1}. $$

Prove ${2n\choose0} + {2n\choose2} + \dots + {2n\choose{2n}} = 2^{2n-1}$

4 Answers4