Fibonacci Formula: Find Nth Term in Sequence

[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.

 

[Santa1]

Look here, http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_Golden_Ratio.

 

[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.

 

[brandy]

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

 

[Defennder]

There are probably other ways to derive it, but I'm only familiar with the one with matrices. There's a current thread on this here:
https://www.physicsforums.com/showthread.php?p=1856158

 

[HallsofIvy]

The terms in a Fibonacci sequence obey the recursive rule F_n+2= F_n+1+ F_n. One common way of solving such equations is to try a solution of the form F_n= aⁿ. Then F_n+1= aⁿ⁺¹ and F_n+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 F_n+2= F_n+1+ F_n.

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.

 

[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.

 

[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.

 

[FeDeX_LaTeX]

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