Sum of Even Index Binomial Coefficients/Proof 2
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Theorem
- $\ds \sum_{i \mathop \ge 0} \binom n {2 i} = 2^{n - 1}$
Proof
Let ${\N_n}^*$ be the initial segment of natural numbers, one-based.
Let:
- $E_n = \set {X : \paren {X \subset {\N_n}^*} \land \paren {2 \divides \size X} }$
- $O_n = \set {X : \paren {X \subset {\N_n}^*} \land \paren {2 \nmid \size X} }$
That is:
- $E_n$ is the set of all subsets of ${\N_n}^*$ with an even number of elements
- $O_n$ is the set of all subsets of ${\N_n}^*$ with an odd number of elements.
So by Cardinality of Set of Subsets:
- $\ds \sum_{i \mathop \ge 0} \binom n {2 i} = 2^{n - 1} \iff \size {E_n} = 2^{n - 1}$
which is to be proved by induction below.
Basis of the Induction
Let $n = 1$.
Then
- $\size {E_n} = \size {\set \O} = 1$
and:
- $2^{n - 1} = 2^{1 - 1} = 2^0 = 1$
This is the basis for the induction.
Induction Hypothesis
This is the induction hypothesis:
- $\size {E_k} = 2^{k - 1}$
So:
- $\size {O_k} = \size {2^{ {\N_k}^*} \divides E_k} = 2^k - 2^{k - 1} = 2^{k - 1}$
Induction Step
This is the induction step:
\(\ds \size {E_{k + 1} }\) | \(=\) | \(\ds \size {\set {X: \paren {X \subset {\N_{k + 1} }^* } \land \paren {2 \divides \size X} } }\) | Definition of $E_n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {\set {X: X \in E_k} \cup \set {X \cup \set {k + 1}: X \in O_k} }\) | Construction of the set with smaller sets, considering the presence of the element $k+1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {E_k} + \size {O_k}\) | $E_k$ and $O_k$ are disjoint by definition | |||||||||||
\(\ds \) | \(=\) | \(\ds 2^{k - 1} + 2^{k - 1}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds 2^k\) |
The result follows by induction.
$\blacksquare$