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
$L_n = a\phi^n + b\widehat{\phi}^n.$

Question

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
$L_n = a\phi^n + b\widehat{\phi}^n.$

ganeshie8 · Answer

try $a=b=1$