The
Fibonacci sequence <Fn>
:0, 1, 1, 2, 3, 5, 8, 13, 21,… which satisfies F0=0, F1=1,
and Fk+2=Fk+1+Fk. The goal of this article
is to find the general form of the kth
term Fk of the Fibonacci sequence.
Let .
Consider
, t
hat is uk+1= Auk , where .
Because the characteristic polynomial of A is, the eigenvalues of A areand, and the corresponding eigenvectors are and
. Of course {x1,
x2} are linearly
independent and generate uk.
Write u0=c1x1+c2x2, that is, and it’s easy to solve the coefficients and . Hence
u1=c1Ax1+c2Ax2= c1λ1x1+c2λ2x2 ,
… ,
uk=c1Akx1+c2Akx2= c1λ1kx1+c2λ2kx2
.
Thus
.