# Fibonacci formula?

1. Aug 31, 2008

### brandy

im just curious. is there a formula for the fibonacci formula in terms of..well terms. like the nth term =..?
iv been trying to figure it out for a couple of days now but am not that smart.

2. Aug 31, 2008

### Santa1

3. Aug 31, 2008

### Defennder

You could derive it, if you know enough elementary linear algebra and in particular diagonalisation of matrices. It's not that difficult. You start off with recursive definition of the n+1 and nth term and n-1 term, put them all into a matrix and show that it is diagonalisable, then write out the matrix equation.

4. Sep 2, 2008

### brandy

done the algebra but have only learnt +-x/ matrices.
how does the n+1 thing work
like i said am not that smart.

5. Sep 2, 2008

6. Sep 2, 2008

### HallsofIvy

The terms in a Fibonacci sequence obey the recursive rule Fn+2= Fn+1+ Fn. One common way of solving such equations is to try a solution of the form Fn= an. Then Fn+1= an+1 and Fn+2= an+2 so the equation becomes an+2= an+1+ an. Dividing by an gives a2= a+ 1 or a2- a- 1= 0. Solving that by the quadratic formula,
$$a= \frac{1\pm\sqrt{5}}{2}$$
In other words,
$$F_n= \left(\frac{1+\sqrt{5}}{2}\right)^n$$
and
$$F_n= \left(\frac{1-\sqrt{5}}{2}\right)^n$$
both satisfy Fn+2= Fn+1+ Fn.

Since that is a linear equation, any solution of that equation can be written
$$A\left(\frac{1+\sqrt{5}}{2}\right)^n+ B\left(\frac{1-\sqrt{5}}{2}\right)^n$$

Now, looking at the first two terms of the Fibonacci sequence
$$F_0= A+ B= 1$$
and
$$F_1=A\left(\frac{1+\sqrt{5}}{2}\right)+ B\left(\frac{1-\sqrt{5}}{2}\right)= 1$$
gives two equations to solve for A and B.

7. Aug 6, 2010

### FeDeX_LaTeX

Sorry for the bump, but could you show me how you would solve for A and B?

I'm not able to solve simultaneous equations in this form;

A + B = 1
Ax + By = 1

Thanks.

8. Aug 6, 2010

### Mentallic

Simply let B=1-A and then substitute this into the second equation, solve for A there and then substitute back into the first to find B.

9. Aug 7, 2010

### FeDeX_LaTeX

Of course! Solving by substitution. Thanks, I forgot about doing that.