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The Fibonacci numbers are defined recursively by the equation \(F_n = F_{n - 1} + F_{n - 2}\)
for every integer \(n \ge 2\), with initial values \(F_0 = 0\) and \(F_1 = 1\). Let \(G_n = F_{3n}\) be every third Fibonacci number. There are constants \(a\) and \(b\) such that every integer \(n \ge 2\) satisfies \(G_n = a G_{n - 1} + b G_{n - 2}\)
Find \((a, b)\).

 

(I put a lot of effort making all the latex properly so please help asap. THX)

 Feb 25, 2021
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Both Melody and CPhill are offline now. I will try to answer that.

 

This means \(F_{3n} = aF_{3n - 3} + bF_{3n - 6}\) by the definition of \(G\).

 

Then, we try some \(F_{3n}, F_{3n - 3}, F_{3n - 6}\). For this to be possible, we need the values of \(F_n\) from n = 0 up to n = 9.

 

\(F_0 = 0\\ F_1 = 1\\ F_2 = 1\\ F_3 = 2\\ F_4 = 3\\ F_5 = 5\\ F_6 = 8\\ F_7 = 13\\ F_8 = 21\\ F_9 = 34\)

 

Then, using the equation on the first line of my solution and substituting n = 2 and n = 3,

 

\(\begin{cases}F_6 = aF_3 + bF_0\\F_9=aF_6+bF_3\end{cases}\)

 

Simplifying,

 

\(\begin{cases}2a = 8\\8a + 2b = 34\end{cases}\)

 

Solving gives \(a = 4, b = 1\).

 

Therefore, \((a, b) = (4, 1)\).

 Feb 25, 2021

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