Origin of the constant $\phi$ in Binet's formula of the $n$-th term of the Fibonacci sequence I have read in a page that " To find the nth term of the Fibonacci series, we can use Binet's Formula"  


F(n) = round( (Phi ^ n) / √5 ) provided n ≥ 0
    where Phi=1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ... .


my question is,  


*

*What is this Phi?

*How they generated this Phi?

 A: $\phi = \frac{1+\sqrt{5}}{2}$ is a very important constant in nature called the golden ratio, it appears in many different guises.
Binets formula is:
$F_n = \frac{(\phi^n - (-\phi)^{-n})}{\sqrt{5}} = \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}}$
There are many different ways to prove Binet's formula. One I like in particular (which generalises to other linear difference equations) is to use eigenvalues and eigenvectors of a particular matrix. This is essentially the reason behind the appearance of $\phi$.
See here:
http://en.wikipedia.org/wiki/Fibonacci_number 
A: The $\varphi$ here is the famous Golden ratio. It appears to be one of those fundamental numbers that arises from the nature itself. The Wikipedia link above provides a good discussion about its nature and its occurences in math.
As for its particular connection to the Fibonacci numbers, Wikipedia again gives a good explanation. (From a philosophical perspective, I would actually say that Fibonacci numbers are more fundamental than $\varphi$, but this is just a personal feeling...)
A: Maybe you can take a look at this : http://en.wikipedia.org/wiki/Golden_ratio.
It is explain how you can calculate the $n^{th}$ term and give a method to calculate $\phi$.
A: Using generating functions:
\begin{align}
\color{brown}{f(x) = \sum_0^\infty F_n x^n} &= F_0x^0 + F_1x^1+F_2 x^2 + F_3 x^3+ F_4 x^4+\cdots\\
xf(x) &=  \quad\quad\quad\,  F_0x^1+F_1x^2+F_2x^3+F_3x^4+\cdots\\
x^2f(x)&=\quad\quad\quad\quad\quad\quad\,\,F_0x^2+F_1x^3 + F_2x^4+\cdots
\end{align}
Since the value of $F_0=0$ and $F_1=F_2=1$, while the recursive formula is $F_{n+2}=F_{n+1}+F_{n}$,
\begin{align}
f(x)  &= F_1x^1+F_2 x^2 + F_3 x^3+ F_4 x^4+\cdots\\
-\big[ xf(x) &=  \quad\quad\quad\,  F_1x^2+F_2x^3+F_3x^4+\cdots\big]\\
-\big[x^2f(x)&=\quad\quad\quad\quad\quad\quad\,\,F_1x^3 + F_2x^4+\cdots\big]\\
\quad\\
&\hline\\
(1-x-x^2)f(x)&=x\\
f(x)&=\frac{x}{1-x-x^2}
\end{align}
The denominator is a quadratic with roots $0=1-x-x^2$; $\large x=\frac{1\pm\sqrt{5}}{-2}$, or $-\phi$ and $-\psi:$
$$\color{red}{\varphi=\frac{1+\sqrt{5}}{2}} \text{ and } \psi=\frac{1-\sqrt{5}}{2}$$ 
Hence,
\begin{align}
(1-x-x^2)f(x)&=x\\
f(x)&=\frac{x}{1-x-x^2}\\
f(x)&=\frac{-x}{(x+\varphi)(x-\psi)}\\
f(x)&=\frac{A}{x+\varphi}+\frac{B}{x-\psi}\tag 1
\end{align}
Therefore,
$$-x=A(x-\psi)+B(x+\varphi)$$
If $x=-\varphi$, $A=-\frac{\varphi}{\sqrt{5}}$, and if $x=-\psi$, $B=\frac{\psi}{\sqrt{5}}.$ Going back to Eq. 1:
$$f(x)=\frac{1}{\sqrt{5}}\left(\color{blue}{\frac{\psi}{x+\psi}}-\color{red}{\frac{\varphi}{x+\varphi}}\right)\tag 2$$
Since $\varphi=-\frac{1}{\psi}$,
$$\color{blue}{\frac{\psi}{x+\psi}}= \frac{1}{1+\frac{x}{\psi}}
=\frac{1}{1-\varphi x}
=\sum_{n=0}^\infty \varphi^n x^n\tag{*}
$$
(*) using the formula for the geometric series, $\sum_0^\infty x^n= \frac{1}{1-x}.$ In the case of
$$\color{red}{\frac{\varphi}{x+\varphi}}=\sum_{n=0}^\infty \psi^n x^n$$
Going back to Eq.2,
$$f(x)=\frac{1}{\sqrt{5}}\left(\sum_{n=0}^\infty \varphi^n x^n-\sum_{n=0}^\infty \psi^n x^n\right)=\sum_0^\infty \frac{1}{\sqrt{5}}\left( \varphi^n - \psi^n \right)x^n=\color{brown}{\sum_0^\infty F_n x^n}$$
So we get Binet's formula:
$$\large F_n=\frac{1}{\sqrt{5}}(\varphi^n-\psi^n)\tag{Binet's formula}$$
