We will show that if v is a solution of the equation x2-x-1=0, then:
vn = Fnv + Fn-1
The assertion is certainly true for n=1 and for n=2, because for n=1 it says v = v and for
n=2 it says v2 = v+1.
In order to prove the correctness of the assertion for n=3 we multiply the equation for n=2
(which we proved to be correct) by v v3 = F2v2 + F1v =
F2(v + 1) + F1v =
(F2 + F1)v + F2 =
F3v + F2.
This shows that the formula is correct for n=3.
We repeat this procedure and multiply the formula for n=3 by v. v4 = F3v2 + F2v =
F3(v + 1) + F2v =
(F3 + F2)v + F3 =
F4v + F3.
This shows that the formula is correct for n=4.
In general: Suppose that the formula is correct for n=N, then the formula is also correct for n=N+1, because: vN+1 = FNv2 + FN-1v =
FN(v + 1) + FN-1v =
(FN + FN-1)v + FN =
FN+1v + FN.
Thus we may conclude that the formula is correct for all n>=0.
r = (1 + 5)/2 en s = (1 - 5)/2 are the two solutions of the equation
x2-x-1=0. This yields: rn = Fnr + Fn-1 and sn = Fns + Fn-1
When we subtract the second of these equations from the first this yields rn-sn = Fn(r-s), or Fn = (rn-sn)/(r-s).
The nth Lucas number Ln is defined by Ln = Fn+1 + Fn-1.
Then Ln = rn + sn.
5)/2 en s = (1 -