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