Can you derive the explicit form of Fibonacci Sequence for arbitrary a0 and a1?

Question

I know the Fibonacci sequence can be written as an explicit form when we assume that a0 and a1 both equal 1. But can this also be done for an arbiary a and b as the values of a0 and a1?

Answer 1

In general we have that the series defined by $a_0=a$, $a_1=b$, $a_n=k(a_{n-1}+a_{n-2})$ has a closed form of $$a_n=\frac{(a\sqrt{k^2+4k}-ak+2b)(k+\sqrt{k^2+4k})^n+(a\sqrt{k^2+4k}+ak-2b)(k-\sqrt{k^2+4k})^n}{2^{n+1}\sqrt{k^2+4k}}$$ So if $k=1$ then this simplifies to $$a_n=\frac{(a\sqrt{5}-a+2b)(1+\sqrt{5})^n+(a\sqrt{5}+a-2b)(1-\sqrt{5})^n}{2^{n+1}\sqrt{5}}$$ $$=\frac{(a\sqrt{5}-a+2b)\left(\frac{1+\sqrt{5}}2\right)^n+(a\sqrt{5}+a-2b)\left(\frac{1-\sqrt{5}}2\right)^n}{2\sqrt{5}}$$

Answer 2

Define a sequence $\left \{ a_{n} \right \}$ defind by $a_{n}=a_{n-2}+a_{n-1} \, ,\, a_{0}=c\, ,\, a_{1}=d.$ then we can write:$$\begin{pmatrix} a_{n+1}\\ a_{n} \end{pmatrix}=\begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}\cdot \begin{pmatrix} a_{n}\\ a_{n-1} \end{pmatrix}=\begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}^2\cdot\begin{pmatrix} a_{n-1}\\ a_{n-2} \end{pmatrix}=...=\begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}^n\cdot\begin{pmatrix} a_{1}\\ a_{0} \end{pmatrix}=\begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}^n\cdot\begin{pmatrix} d\\ c \end{pmatrix}.$$ We got a formula for $a_{n}$ based on $c$ and $d$.