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

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

$\blacksquare$


Also see


Sources

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