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Question: Show that \(r\, = \,\dfrac{{1 + \sqrt 5 }}{2}\)for the Fibonacci sequence?...

Show that r = 1+52r\, = \,\dfrac{{1 + \sqrt 5 }}{2}for the Fibonacci sequence?

Explanation

Solution

This value of r = 1+52r\, = \,\dfrac{{1 + \sqrt 5 }}{2}is called the golden ratio or golden number. We have many applications of golden ratio in nature. Sequence of number 0,1,1,2,3,5,8,13,… is called Fibonacci Sequence and the number of this series is Fibonacci Number. To get this value of rr we will use methods of root of a quadratic equation and definition of Fibonacci sequence.

Complete step by step solution:
Lets us see, how do we getr = 1+52r\, = \,\dfrac{{1 + \sqrt 5 }}{2} from the
Fibonacci sequence?

Before proving this we will write the recurrence relation of the Fibonacci sequence and then convert it to a quadratic equation. Solve the root of the quadratic equation using the quadratic formula to getr = 1+52r\, = \,\dfrac{{1 + \sqrt 5 }}{2}.

Definition: Fibonacci series is defined recursively by this recurrence relation.
$
{F_0} = 0 \\

{F_1} = 1 \\

{F_n} = {F_{n - 1}} + {F_{n - 2}},,,,,,,,,,,,,,(\forall ,n = 2) \\
$

If we suppose that the general term of Fibonacci sequence isFn=arn{F_n} = a{r^n}. So our recurrence relation of Fibonacci sequence will become
arn = arn−1+arn−2            (∀ n=2)a{r^n}\, = \,a{r^{n - 1}} + a{r^{n - 2}}\,\,\,\,\,\,\,\,\,\,\,\,(\forall \,n = 2)

Taking out arn−2a{r^{n - 2}}from both side of the above equation we get
r2 = r + 1{r^2}\, = \,r\, + \,1

r2−r−1 = 0{r^2} - r - 1\, = \,0

Now, applying the quadratic formula we find the root of the quadratic equationr2−r−1 = 0{r^2} - r - 1\, = \,0. We get roots as follows:

$
a = 1 \\

b = - 1 \\

c = - 1 \\
$

$
r = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

r = \dfrac{{1 \pm \sqrt {{{( - 1)}^2} - 4 \cdot 1 \cdot ( - 1)} }}{{2 \cdot 1}} \\

r = \dfrac{{1 \pm \sqrt {1 + 4} }}{2} \\

r = \dfrac{{1 \pm \sqrt 5 }}{2} \\
$

Hence we get the root of quadratic equation r2−r−1 = 0{r^2} - r - 1\, = \,0which we got from the recurrence relation of the Fibonacci sequence is r=1+52r = \dfrac{{1 + \sqrt 5 }}{2} andr=1−52r = \dfrac{{1 - \sqrt 5 }}{2}.

The number r=1+52r = \dfrac{{1 + \sqrt 5 }}{2} is called the golden ratio or golden number.

Note: General term of Fibonacci sequence is of the form Fn = A(1+52) +B(1−52){F_n}\, = \,A\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\, + B\left( {\dfrac{{1 - \sqrt 5 }}{2}} \right) and we get the value of constant A and B from initial condition of recurrence of the Fibonacci series which isF0=0    and    F1=1{F_0} = 0\,\,\,\,and\,\,\,\,{F_1} = 1. We can also find the roots of quadratic also using the completing square and factor method. Similarly, we can also find the general term of Lucas series.