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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: Exercise $25$