# OZO: Urgent

Let r satisfy the equation r^2 = r + 1.
Show that the sequence s_n = ar^n, where A is constant, satisfies the fibonacci equation f_n = f_(n-1) + f_(n-2).

I'm not sure how to show this.

I know that f_1 = 1 & f_2 = 2

f_1 = 1
f_2 = 2
f_3 = f_2 + f_1
f_4 = f_3 + f_2

So

s_3 = s_2 + s_1 = Ar + Ar
s_4 = s_3 + s_2 = Ar + Ar + Ar
s_5 = s_4 + s_3 = Ar + Ar + Ar + Ar + Ar
s_6 = s_5 + s_4 = Ar + Ar + Ar + Ar + Ar + Ar + Ar + Ar

I'm not sure what i'm doing ;/

-Brian
LVL 19
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Commented:
Hi BrianGEFF719,

May be I can try this one ... It can be proved using induction

s_n = ar^n
so ...
s_0 = a
s_1 = ar
s_2 = ar^2  .... a+ar = a(1+r) = ar^2

similarly
s_k = ar^k .... ar^(k-1) + ar^(k-2) = (ar^(k-2)) * (1+r) = (ar^(k-2) ) * r^2 = ar^k

Cheers!
Sunnycoder

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Commented:
f_(k) = f_(k-1) + f_(k-2) = ar^(k-1) + ar^(k-2) = ar^(k-2) * (r+1) = ar^(k-2) * r^2 = ar^k
Commented:
But I guess sunnycoder already said that.
Author Commented:
Hey Sunnycoder,

About 2 1/2yrs ago you used to help me with algebra...I've since advanced to higher levels of math, but good to see you again :)

Brian
Author Commented:
Ozo, if you've got a few minutes i've got an easy physics one for you...
Author Commented:
Author Commented:
I dont quite understand this step:

ar^(k-1) + ar^(k-2) = ar^(k-2)) * (1+r)

How do you guys get this?
Commented:
ar^(k-2) is common in both terms
Author Commented:
Sunnycoder, thanks a lot!
Commented:
:)
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