Fibonacci Number by Power of 2/Proof 2

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Theorem

\(\ds \forall n \in \Z_{\ge 0}: \, \) \(\ds 2^{n - 1} F_n\) \(=\) \(\ds \sum_k 5^k \dbinom n {2 k + 1}\)
\(\ds \) \(=\) \(\ds \dbinom n 1 + 5 \dbinom n 3 + 5^2 \dbinom n 5 + \cdots\)

where:

$F_n$ denotes the $n$th Fibonacci number
$\dbinom n {2 k + 1} \ $ denotes a binomial coefficient.


Proof

\(\ds 2^{n - 1} F_n\) \(=\) \(\ds \dfrac {2^n} {2 \sqrt 5} \paren {\phi^n - \hat \phi^n}\) Euler-Binet Formula
\(\ds \) \(=\) \(\ds \dfrac {\paren {1 + \sqrt 5}^n - \paren {1 - \sqrt 5}^n} {2 \sqrt 5}\) Definition 2 of Golden Mean
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \sqrt 5} \sum_{j \mathop = 0}^n \binom n j \sqrt 5^j - \dfrac 1 {2 \sqrt 5} \sum_{j \mathop = 0}^n \paren {-1}^j \binom n j \sqrt 5^j\) Binomial Theorem
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt 5} \sum_{\substack {0 \mathop \le j \mathop \le n \\ j \text { odd} } } \binom n j \sqrt 5^j\) even terms vanish, odd terms double up
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt 5} \sum_{j \text { odd} } \binom n j 5^{j / 2}\) Definition of Binomial Coefficient: $\dbinom n j = 0$ for $j < 0$ and $j > n$
\(\ds \) \(=\) \(\ds \sum_{j \text { odd} } \binom n j 5^{\paren {j - 1} / 2}\) gathering the spare $\sqrt 5$ into the index


Setting $j = 2 k + 1$ for $0 \le k \le \paren {j - 1} / 2$ gives:

$\ds \sum_{k \mathop \ge 0} \binom n {2 k + 1} 5^k$

and the limits of the index of the summation are irrelevant, as $\dbinom n {2 k + 1} = 0$ for $j < 0$ and $j > n$.

Hence the result.


Historical Note

This result was discovered by Eugène Charles Catalan.


Sources

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