Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less

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Theorem

Let $n \in \Z$.

Then:

$\hat \phi^n = F_n \hat \phi + F_{n - 1}$

where:

$F_n$ denotes the $n$th Fibonacci number
$\hat \phi$ denotes the $1$ minus the golden mean:
$\hat \phi := 1 - \phi$


Proof

\(\ds F_n \hat \phi + F_{n - 1}\) \(=\) \(\ds F_n \paren {-\dfrac 1 \phi} + F_{n - 1}\) Reciprocal Form of One Minus Golden Mean
\(\ds \) \(=\) \(\ds -\dfrac 1 \phi \paren {F_n - \phi F_{n - 1} }\)
\(\ds \) \(=\) \(\ds -\dfrac 1 \phi \paren {\paren {-1}^{n + 1} F_{-n} - \phi \paren {-1}^n F_{-\paren {n - 1} } }\) Fibonacci Number with Negative Index
\(\ds \) \(=\) \(\ds -\dfrac 1 \phi \paren {\paren {-1}^{n + 1} F_{-n} + \phi \paren {-1}^{n + 1} F_{-\paren {n - 1} } }\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \paren {-\dfrac 1 \phi \paren {F_{-n} + \phi F_{-n + 1} } }\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \paren {-\dfrac 1 \phi } \paren {\phi^{-n + 1} }\) Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less
\(\ds \) \(=\) \(\ds \paren {-1}^n \paren {\dfrac 1 {\phi^n} }\)
\(\ds \) \(=\) \(\ds \paren {-\dfrac 1 \phi}^n\)
\(\ds \) \(=\) \(\ds \hat \phi^n\)

$\blacksquare$


Also see


Sources

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