Fibonacci sequence

The Fibonacci sequence is a recursive sequence, defined as
 * $$a_0=0, a_1=1$$
 * $$\{a_i\}_{k=0}^{\infty} = a_{i-1} + a_{i-2} = 0, 1, 1, 2, 3, 5, 8, 13, 21, ...$$

Properties

 * $$\lim_{n \to \infty} \frac{a_{n + 1}}{a_n} = \phi$$

$$\phi$$ is equal to the golden ratio.

$$a_n = \frac{\phi^n - (1-\phi)^n}{\sqrt 5 } = \frac{(1+\sqrt 5)^n - (1-\sqrt 5)^n}{ 2^n \sqrt 5 } $$

Proof
We are given this recurrence relation,
 * $$F_{n} = F_{n-1}+F_{n-2}$$

Which is subject to

$$F_{1} = 1$$ and

$$F_{2} = 1$$ . One may form an auxiliary equation in

$$\lambda$$ accordingly and solve for

$$\lambda$$ .


 * $$\lambda^2 - \lambda - 1 = 0$$

Through the use of the quadratic formula, one will obtain,


 * $$\lambda = \frac{1+\sqrt{5}}{2}$$or$$\lambda = \frac{1-\sqrt{5}}{2}$$

Or otherwise,


 * $$\lambda = \phi$$or$$\lambda = 1 - \phi$$

So we have,


 * $$F_{n} = C_1(\phi)^n + C_2(1-\phi)^n$$

Where

$$C_1$$ and

$$C_2$$ are constants to be determined. Substituting the values we have


 * $$1 = C_1\phi + C_2(1-\phi)$$
 * $$1 = C_1\phi^2 + C_2(1-\phi)^2$$

Solving for both variables, we obtain,


 * $$C_1 = \frac{1}{\sqrt{5}}$$
 * $$C_2 = -\frac{1}{\sqrt{5}}$$

So, one has


 * $$F_n = \frac{1}{\sqrt{5}} \left(\phi^n - [1-\phi]^n\right)$$

As required

$$\blacksquare$$

Sum
For all integers

$$n$$ greater than or equal to

$$1$$ ,


 * $$\sum_{t=1}^{n} F_{t} = F_{n+2} - 1$$

Proof
Proposition: given

$$F_{n}$$ as defined above,
 * $$\sum_{t=1}^{n} F_{t} = F_{n+2} - 1$$$$\forall n \in \mathbb{N}$$

Let

$$n = 1$$ ,


 * $$\sum_{t=1}^{n} F_{t} = F_{1} = 1 = F_{3} - 1$$

Therefore the proposition holds for

$$n=1$$ . Assume that the proposition holds for

$$n=\lambda$$ . We may now make use of the inductive step. Let

$$n=\lambda + 1$$ .


 * $$\sum_{t=1}^{\lambda+1} F_{t} = \sum_{t=1}^{\lambda} F_{t} + F_{\lambda + 1}$$.

We know that

$$\sum_{t=1}^{\lambda} F_{t} = F_{\lambda + 2} - 1$$ , from the assumption that the proposition holds for

$$n = \lambda$$ .

So, we have,
 * $$\sum_{t=1}^{\lambda+1} F_{t} = F_{\lambda + 2} + F_{\lambda + 1} - 1$$

Using the definition,


 * $$F_{n} = F_{n-1} + F_{n-2}$$

One obtains


 * $$\sum_{t=1}^{\lambda+1} F_{t} = F_{\lambda + 3} - 1$$

Which obeys the proposition


 * $$\sum_{t=1}^{\lambda} F_{t} = F_{\lambda + 2} - 1$$.

As the proposition holds for

$$n=1$$ ,

$$n = \lambda$$ and

$$n = S(\lambda)$$ , the proposition holds for all natural numbers.

$$\blacksquare$$

Binet's Formula
Binet's Formula is a theorem that allows one to determine

$$ F_n $$ , where

$$ F_n $$ represents the

$$ n^{th} $$ Fibonacci Number.

The theorem is as follows:

$$ F_n= \frac{1}{\sqrt{5}} \cdot ((\frac{1+\sqrt{5}}{2})^n - (\frac{1-\sqrt{5}}{2})^n). $$

Trivia

 * Fibonacci numbers are claimed to be common in nature; for example, the shell of a nautilus being a Fibonacci spiral. However, this has been disputed with the spiral having a ratio measured between 1.24 to 1.43.

Terms
Fibonacci-Folge