Euler-Binet Formula

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

where $$\phi$$ is the golden mean.

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

Proof
Proof by induction:

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


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

Basis for the Induction

 * $$P(1)$$ is the case $$\frac {\phi - \hat \phi} {\sqrt 5} = 1 = F \left({1}\right)$$ (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:
 * $$F \left({k}\right) = \frac {\phi^k - \hat \phi^k} {\sqrt 5}$$.

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

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

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:
 * $$U_n = \frac {\alpha^n - \beta^n} {\Delta}$$

where:

$$ $$ $$

where $$m=1$$.

He 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.