Fibonacci Number 2n equals Fibonacci Number n by Lucas Number n
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Theorem
Let $F_n$ denote the $n$th Fibonacci number.
Let $L_n$ denote the $n$th Lucas number.
Then:
- $F_{2 n} = F_n L_n$
Proof
Let:
- $\phi = \dfrac {1 + \sqrt 5} 2$
- $\hat \phi = \dfrac {1 - \sqrt 5} 2$
Then:
\(\ds F_{2 n}\) | \(=\) | \(\ds \dfrac {\phi^{2 n} - \hat \phi^{2 n} } {\sqrt 5}\) | Euler-Binet Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {\phi^n + \hat \phi^n} \paren {\phi^n - \hat \phi^n} } {\sqrt 5}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds L_n \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}\) | Closed Form for Lucas Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds L_n F_n\) | Euler-Binet Formula |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $11$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $11$