Fibonacci Number n+1 Minus Golden Mean by Fibonacci Number n/Proof 2
Jump to navigation
Jump to search
Theorem
- $F_{n + 1} - \phi F_n = \hat \phi^n$
where:
- $F_n$ denotes the $n$th Fibonacci number
- $\phi$ denotes the golden mean.
Proof
\(\ds F_n\) | \(=\) | \(\ds \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}\) | Euler-Binet Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^n - \hat \phi^n} {\frac {1 + \sqrt 5} 2 - \frac {1 - \sqrt 5} 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^n - \hat \phi^n} {\phi - \hat \phi}\) | Definition 2 of Golden Mean |
Thus from Recurrence Relation where n+1th Term is A by nth term + B to the n we have:
- $F_{n + 1} = \phi F_n + \hat \phi^n$
whence the result.
$\blacksquare$