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FibonacciBlocks

Geometric representation of the Fibonacci numbers

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 $

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.[1]

References

  1. Peterson, Ivars (April 1, 2005). "Sea Shell Spirals". Science News. https://www.sciencenews.org/article/sea-shell-spirals. 

Terms

1 1
2 1
3 2
4 3
5 5
6 8
7 13
8 21
9 34
10 55
11 89
12 144
13 233
14 377
15 610