- #1
alec_tronn
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Homework Statement
We were asked in a class discussion if we could try to figure out the recurrence relation of the Fibonacci squared sequence. I correctly figured that the equation was
F[tex]^{2}_{n}[/tex]=2F[tex]^{2}_{n-1}[/tex] + 2F[tex]^{2}_{n-2}[/tex] - F[tex]^{2}_{n-3}[/tex]
with F[tex]^{2}_{1}[/tex] = 0 and F[tex]^{2}_{2}[/tex] = 1 and F[tex]^{2}_{3}[/tex] = 1.
Now, our class homework is to prove it.
The Attempt at a Solution
My first instinct is proof by induction. So I start with:
Test the hypothesis with n=4 (since n=1,2,3 are given by definition).
F[tex]^{2}_{4}[/tex] = 2F[tex]^{2}_{3}[/tex] + 2F[tex]^{2}_{2}[/tex] - F[tex]^{2}_{1}[/tex]
= 2 + 2 - 0
= 4, which agrees with original definition(because F[tex]^{1}_{4}[/tex] = 2, and thus F[tex]^{2}_{4}[/tex] = 4).
So, the formula holds for some arbitrary but fixed n (n being a member of the natural numbers).
Test n+1.
F[tex]^{2}_{n+1}[/tex] = (F[tex]^{1}_{n}[/tex] + F[tex]^{1}_{n-1}[/tex])^{2} (by definition. of Fib. Sequence)
=F[tex]^{2}_{n}[/tex] + 2 F[tex]^{1}_{n}[/tex] * F[tex]^{1}_{n-1}[/tex] + F[tex]^{2}_{n-1}[/tex] (by distribution)
= 2F[tex]^{2}_{n-1}[/tex] + 2F[tex]^{2}_{n-2}[/tex] - F[tex]^{2}_{n-3}[/tex] + 2 F[tex]^{1}_{n}[/tex] * F[tex]^{1}_{n-1}[/tex] + 2F[tex]^{2}_{n-2}[/tex] + 2F[tex]^{2}_{n-3}[/tex] - F[tex]^{2}_{n-4}[/tex] (by the induction hypothesis)
= 2F[tex]^{1}_{n-1}[/tex] * (F[tex]^{1}_{n}[/tex] + F[tex]^{1}_{n-1}[/tex]) + 4F[tex]^{2}_{n-2}[/tex] + F[tex]^{2}_{n-3}[/tex] - F[tex]^{2}_{n-4}[/tex]
=2F[tex]^{1}_{n-1}[/tex] * F[tex]^{1}_{n+1}[/tex] + 4F[tex]^{2}_{n-2}[/tex] + F[tex]^{2}_{n-3}[/tex] - F[tex]^{2}_{n-4}[/tex]
That's where I'm stuck. I know (or think I know) that I'm supposed to end up with :
F[tex]^{2}_{n+1}[/tex] = 2F[tex]^{2}_{n}[/tex] + 2F[tex]^{2}_{n-1}[/tex] - F[tex]^{2}_{n-2}[/tex] and so for every n for which the formula is valid, the formula is also valid for n+1. By induction, the formula works for all of the natural numbers.
Could anyone help me figure out the part I'm stuck at?