Fibonacci sequence

File:FibonacciBlocks.svg
The **Fibonacci sequence** is a recursive sequence, defined as

a
0

=
0
,

a
1

=
1

{\displaystyle a\_0=0, a\_1=1}

{

a
i

}

# k
0

∞

=

a

i
−
1

a

i
−
2

=
0
,
1
,
1
,
2
,
3
,
5
,
8
,
13
,
21
,
.
.
.

{\displaystyle \{a\_i\}\_{k=0}^{\infty} = a\_{i-1} + a\_{i-2} = 0, 1, 1, 2, 3, 5, 8, 13, 21, ...}

## Properties

lim

n
→
∞

a

n
1

a
n

=
ϕ

{\displaystyle \lim\_{n \to \infty} \frac{a\_{n + 1}}{a\_n} = \phi}

ϕ

{\displaystyle \phi}

is equal to the golden ratio.

a
n

=

ϕ
n

−
(
1
−
ϕ

)
n

5

=

(
1

5

)
n

−
(
1
−

5

)
n

2
n

5

{\displaystyle 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

{\displaystyle F\_{n} = F\_{n-1} + F\_{n-2} }

Which is subject to

F
1

=
1

{\displaystyle F\_1=1}

and

F
2

=
1

{\displaystyle F\_2=1}

. One may form an auxiliary equation in

λ

{\displaystyle \lambda}

accordingly and solve for

λ

{\displaystyle \lambda}

.

λ
2

−
λ
−
# 1
0

{\displaystyle \lambda^2 - \lambda - 1 = 0}

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

# λ

1

5

2

{\displaystyle \lambda = \frac{1+\sqrt{5}}{2}}

or

# λ

1
−

5

2

{\displaystyle \lambda = \frac{1-\sqrt{5}}{2}}

Or otherwise,

# λ
ϕ

{\displaystyle \lambda = \phi}

or

# λ
1
−
ϕ

{\displaystyle \lambda = 1 - \phi}

So we have,

F

n

=

C
1

(
ϕ

)
n

C
2

(
1
−
ϕ

)
n

{\displaystyle F\_{n} = C\_1(\phi)^n + C\_2(1-\phi)^n}

Where

C
1

{\displaystyle C\_1}

and

C
2

{\displaystyle C\_2}

are constants to be determined. Substituting the values we have

# 1

C
1

ϕ

C
2

(
1
−
ϕ
)

{\displaystyle 1 = C\_1\phi + C\_2(1-\phi)}

# 1

C
1

ϕ
2

C
2

(
1
−
ϕ

)
2

{\displaystyle 1 = C\_1\phi^2 + C\_2(1-\phi)^2}

Solving for both variables, we obtain,

C
1

=

1

5

{\displaystyle C\_1 = \frac{1}{\sqrt{5}}}

C
2

=
−

1

5

{\displaystyle C\_2 = -\frac{1}{\sqrt{5}}}

So, one has

F
n

=

1

5

(

ϕ
n

−
[
1
−
ϕ

]
n

)

{\displaystyle F\_n = \frac{1}{\sqrt{5}} \left(\phi^n - [1-\phi]^n\right)}

As required

◼

{\displaystyle \blacksquare }

## Sum

For all integers

n

{\displaystyle n}

greater than or equal to

1

{\displaystyle 1}

,

∑

# t
1

n

F

t

=

F

n
2

−
1

{\displaystyle \sum\_{t=1}^{n} F\_{t} = F\_{n+2} - 1}

### Proof

**Proposition**: given

F
n

{\displaystyle F\_n}

as defined above,

∑

# t
1

n

F

t

=

F

n
2

−
1

{\displaystyle \sum\_{t=1}^{n} F\_{t} = F\_{n+2} - 1}

∀
n
∈

N

{\displaystyle \forall n\in \N}

Let

# n
1

{\displaystyle n=1}

,

∑

# t
1

n

F

t

=

F

1

=
# 1

F

3

−
1

{\displaystyle \sum\_{t=1}^{n} F\_{t} = F\_{1} = 1 = F\_{3} - 1}

Therefore the proposition holds for

# n
1

{\displaystyle n=1}

. Assume that the proposition holds for

# n
λ

{\displaystyle n = \lambda}

. We may now make use of the inductive step. Let

# n
λ
1

{\displaystyle n=\lambda + 1}

.

∑

# t
1

λ
1

F

t

=

∑

# t
1

λ

F

t

F

λ
1

{\displaystyle \sum\_{t=1}^{\lambda+1} F\_{t} = \sum\_{t=1}^{\lambda} F\_{t} + F\_{\lambda + 1}}

.
We know that

∑

# t
1

λ

F

t

=

F

λ
2

−
1

{\displaystyle \sum\_{t=1}^{\lambda} F\_{t} = F\_{\lambda + 2} - 1}

, from the assumption that the proposition holds for

# n
λ

{\displaystyle n = \lambda}

.

So, we have,

∑

# t
1

λ
1

F

t

=

F

λ
2

F

λ
1

−
1

{\displaystyle \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

{\displaystyle F\_{n} = F\_{n-1} + F\_{n-2} }

One obtains

∑

# t
1

λ
1

F

t

=

F

λ
3

−
1

{\displaystyle \sum\_{t=1}^{\lambda+1} F\_{t} = F\_{\lambda + 3} - 1}

Which obeys the proposition

∑

# t
1

λ

F

t

=

F

λ
2

−
1

{\displaystyle \sum\_{t=1}^{\lambda} F\_{t} = F\_{\lambda + 2} - 1}

.
As the proposition holds for

# n
1

{\displaystyle n=1}

,

# n
λ

{\displaystyle n = \lambda}

and

# n
S
(
λ
)

{\displaystyle n = S(\lambda)}

, the proposition holds for all natural numbers.

◼

{\displaystyle \blacksquare }

## Binet's Formula

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

F
n

{\displaystyle F\_n}

, where

F
n

{\displaystyle F\_n}

represents the

n

t
h

{\displaystyle n^{th}}

**Fibonacci Number.**

The theorem is as follows:

F
n

=

1

5

⋅
(
(

1

5

2

)
n

−
(

1
−

5

2

)
n

)
.

{\displaystyle 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

The first 15 terms of the Fibonacci sequence are: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181