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# fibonaci sequence question i need help fast pls @Cphill @Melody

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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 $$(a, b)$$.

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