When I learned sums and sequences in algebra II with trig I learned about recursive rules and explicit rules.
A recursive rule written with the formula of: $$a_n = r * a_{n-1}$$ Or as: $$a_n = a_{n-1} + d$$ And an explicit rule written with the formula of: $$a_n = a_1 + (n – 1)d$$ Or as: $$a_n = a_1 * r^{n-1}$$ My math teacher told me that every recursive rule can be written as an explicit rule too and I found that to hold true through all of the math problems I did for homework. However, when I thought of the Fibonacci sequence, $$a_n = a_{n-1} + a_{n-2};\ \ \ a_0 = 0,\ a_1 = 1$$ I couldn't figure out how to turn it into an explicit formula. Is there any way to do this with just the mathematical skills of a pre-calculus student?
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$\begingroup$You could use Binet's formula: $$F_n=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\sqrt{5}}$$
A good derivation is given here, and it should be easily accessible to a pre-calculus student.
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