Sooo everything in our notes about this doesnt really let me know how to complete this proof but...here is what I got.

Question: Use the PCI to prove the following properties of Fibonacci Number: (The stuff after the F is F sub that.)
F(n+6) = 4F(n+3) + F(n) for all natural numbers n.

Proof: Let {F(n): n exist in the Natural numers} be the Fibonnaci sequence.
i) In the special case where n=1 and n=2, we see that:
F(1+6)=F(7)=13=4F(4) + F(1) = 4(3) + 1 = 13
F(2+6)=F(8)=21=4F(5) + F(2) = 4(5) + 1 = 21

ii) Suppose for all n t hat exists in {1,2,...., m-1} F(n+6) = 4F(n+3) + F(n).
F(m) = 4F(m-3) + F(m-6) = ??????????


Thats how far I got. Our examples have something to sub in but im not seeing it. Help please

3/22/2007 10:22:20 PM

im thinking something like, is m-1 is in that series, then m-3 is and so is m-6 but dont know what to write instead of it or replace it with

3/22/2007 10:28:52 PM