Fibonacci Number 3n in terms of Fibonacci Number n and Lucas Number 2n
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Theorem
Let $F_n$ denote the $n$th Fibonacci number.
Let $L_n$ denote the $n$th Lucas number.
Then:
- $F_{3 n} = F_n \paren {L_{2 n} + \paren {-1}^n}$
Proof
Let:
- $\phi = \dfrac {1 + \sqrt 5} 2$
- $\hat \phi = \dfrac {1 - \sqrt 5} 2$
Then:
\(\ds F_{3 n}\) | \(=\) | \(\ds \dfrac {\phi^{3 n} - \hat \phi^{3 n} } {\sqrt 5}\) | Euler-Binet Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {\phi^n - \hat \phi^n} \paren {\phi^{2 n} + \phi^n \hat \phi^n + \hat \phi^{2 n} } } {\sqrt 5}\) | Difference of Two Cubes | |||||||||||
\(\ds \) | \(=\) | \(\ds F_n \paren {\phi^{2 n} + \phi^n \hat \phi^n + \hat \phi^{2 n} }\) | Euler-Binet Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds F_n \paren {L_{2 n} + \phi^n \hat \phi^n}\) | Closed Form for Lucas Numbers |
Then we note:
\(\ds \phi \hat \phi\) | \(=\) | \(\ds \dfrac {1 + \sqrt 5} 2 \dfrac {1 - \sqrt 5} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 - 5} 4\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds -1\) |
The result follows.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $11$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $11$