The Fibonacci numbers are defined recursively by the equation Fn=Fn−1+Fn−2
for every integer n≥2, with initial values F0=0 and F1=1. Let Gn=F3n be every third Fibonacci number. There are constants a and b such that every integer n≥2 satisfies Gn=aGn−1+bGn−2
Find (a,b).
(I put a lot of effort making all the latex properly so please help asap. THX)
+9675 Both Melody and CPhill are offline now. I will try to answer that.
This means F3n=aF3n−3+bF3n−6 by the definition of G.
Then, we try some F3n,F3n−3,F3n−6. For this to be possible, we need the values of Fn from n = 0 up to n = 9.
F0=0F1=1F2=1F3=2F4=3F5=5F6=8F7=13F8=21F9=34
Then, using the equation on the first line of my solution and substituting n = 2 and n = 3,
{F6=aF3+bF0F9=aF6+bF3
Simplifying,
{2a=88a+2b=34
Solving gives a=4,b=1.
Therefore, (a,b)=(4,1).
