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

$\blacksquare$


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