## zmudz one year ago The Fibonacci sequence, with $$F_0 = 0$$, $$F_1 = 1$$ and $$F_n = F_{n - 2} + F_{n - 1$$, had a closed form $$F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right),$$ where $$\phi = \frac{1 + \sqrt{5}}{2} \; \text{and} \; \widehat{\phi} = \frac{1 - \sqrt{5}}{2}.$$ The Lucas numbers are defined in a similar way. Let $$L_0$$ be the zeroth Lucas number and $$L_1$$ be the first. If \begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n - 1} + L_{n - 2} \; \text{for}\; n \geq 2 \end{align*} Find $$(a,b)$$ such that $$L_n = a\phi^n + b\widehat{\phi}^n.$$

try $$a=b=1$$