General solution expressed in $a_0$ and $a_1$ of a Fibonacci-like sequence?

Question

What is the general solution expressed in $a_0$ and $a_1$ of a Fibonacci-like sequence ?

I mean if $a_0,a_1$ are given and $a_{n+1}:=a_n+a_{n-1}$

$(\begin{array}{cc}a_n&a_{n-1}\end{array})=(\begin{array}{cc}a_{n-1}&a_{n-2}\end{array})\left(\begin{array}{cc}1&1\\1&0\end{array}\right)=(\begin{array}{cc}a_{1}&a_{0}\end{array})\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^{n-1}$

or according to this formula; $a_n=a_1\sum\limits_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n-k-1}{k}+a_0\sum\limits_{k=0}^{\left\lfloor\frac{n-2}{2}\right\rfloor}\binom{n-k-2}{k}$

Answer 1

The general solution will be as follows: $a_n = \dfrac{\varphi_2a_0 - a_1}{\varphi_2 - \varphi_1}\varphi_1^{n} + \dfrac{a_1 - \varphi_1a_0}{\varphi_2 - \varphi_1}\varphi_2^{n}$, where $\varphi_{1,2} = \dfrac{1 \pm \sqrt{5}}{2}$ -- roots of equation $\varphi^2-\varphi-1 = 0$. You can check it by substituting this formula to the recurrence.

Answer 2

We can solve the recurrence to get

$$a_n=\frac{(a_0\chi-a_1)\phi^n-(a_0\phi-a_1)\chi^n}{\chi-\phi}$$

Where $\phi$ is the positive solution of $x^2=x+1$ and $\chi$ is the negative.