Euler-Binet Formula

Theorem
The Fibonacci numbers have a closed-form solution:
 * $\displaystyle F \left({n}\right) = \frac {\phi^n - \left({1 - \phi}\right)^n} {\sqrt 5} = \frac {\phi^n - \left({-1 / \phi}\right)^n} {\sqrt 5}$

where $\phi$ is the golden mean.

Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
 * $\displaystyle F \left({n}\right) = \frac {\phi^n - \hat \phi^n} {\sqrt 5}$

Proof
Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle F \left({n}\right) = \frac {\phi^n - \hat \phi^n} {\sqrt 5}$

$P(0)$ is true, as this just says:
 * $\displaystyle \frac {\phi^0 - \hat \phi^0} {\sqrt 5} = \frac {1 - 1} {\sqrt 5} = 0 = F(0)$

Basis for the Induction
$P(1)$ is the case:
 * $\dfrac {\phi - \hat \phi} {\sqrt 5} = 1 = F \left({1}\right)$

(after some algebra).

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle F \left({k}\right) = \frac {\phi^k - \hat \phi^k} {\sqrt 5}$

Then we need to show:
 * $\displaystyle F \left({k+1}\right) = \frac {\phi^{k+1} - \hat \phi^{k+1}} {\sqrt 5}$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N: F \left({n}\right) = \frac {\phi^n - \hat \phi^n} {\sqrt 5}$

Alternative Proof
This follows as a direct application of the first Binet form:


 * $U_n = m U_{n-1} + U_{n-2}$

where:

has the closed-form solution:
 * $U_n = \dfrac {\alpha^n - \beta^n} {\Delta}$

where:

where $m=1$.

It is also known as Binet's Formula.

Binet derived it in 1843, but it was already known to Euler, de Moivre and Daniel Bernoulli over a century earlier.

However, it was Binet who derived the more general Binet Form of which this is an elementary application.