## 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*}

1. zmudz

Let \begin{align*} A_0 &= 6 \\ A_1 &= 5 \\ A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2. \end{align*} There is a unique ordered pair $$(c,d)$$ such that $$c\phi^n + d\widehat{\phi}^n$$ is the closed form for sequence $$A_n$$. Find $$c$$ using the Fibonacci and Lucas number sequences.

2. ganeshie8

It seems $\large A_n=c\phi^n+d\widehat{\phi}^n = 2F_n+3L_n$

3. anonymous

I don't know why the problem is arranged too complicated. To my course, it is easy to find it out by characteristics equation $$r^2-r-1=0$$, which gives us $$r_1= \dfrac{1+\sqrt5}{2},r_2=\dfrac{1-\sqrt5}{2}$$, Hence the closed form of $$A_n =C(\dfrac{1+\sqrt 5}{2})^n+D(\dfrac{1-\sqrt5}{2})^n$$ If $$A_0=6\rightarrow C+D=6$$ If$$A_1= 5\rightarrow C(\dfrac{1+\sqrt5}{2}) +D(\dfrac{1-\sqrt5}{2})=5$$ Solve for C, D, we have $$C = 9-\dfrac{2}{\sqrt5}, D= \dfrac{2}{\sqrt5}-3$$

4. dumbcow

just a correction but i believe the solution to the above system is: $C = 3 + \frac{2}{\sqrt{5}}$ $D = 3 - \frac{2}{\sqrt{5}}$