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.

My Solution:

$A_n$ is just $3L_n$+$2F_n,$ where $L_n$ is the $n^{th}$ Lucas number and $F_n$ is the $n^{th}$ Fibonacci number. Since the closed form of the Fibonacci sequence is $$F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right),$$ and the closed form of the Lucas sequence is $$L_n = \phi^n + \widehat{\phi}^n,$$ we get the closed form of $A_n$ as $\dfrac{2}{\sqrt{5}}+3.$

What is wrong with my solution?

EDIT: My answer is actually right.

  • $\begingroup$ Simply, that is not the closed form of $A_n$, but the closed form of $c$. $\endgroup$ – Jack D'Aurizio Apr 2 '17 at 21:30
  • $\begingroup$ See en.wikipedia.org/wiki/…. $\endgroup$ – lhf Apr 2 '17 at 21:48

Any sequence of the Fibonacci-type, i.e., $f_n=f_{n-1}+f_{n-2}$, can be expressed as a sum of a Fibonacci-term and a Lucas-term, as follows

$$f_n=\left(f_1-\frac{f_0}{2}\right) \frac{\phi^n-\psi^n}{\phi-\psi}+\frac{f_0}{2} (\phi^n+\psi^n) $$

where $\psi=-1/\phi$. For a more detailed description and generalization, see my post at Decimal Fibonacci Number?


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