I'm reading the Linear Algebra Solutions Manual and there are some things I can't quite get around recurrently defined determinants.
We have a determinant which is also a recurrent relation $\Delta_n =p\Delta_{n-1} + q\Delta_{n-2}$
We represent $p$ and $q$ in the form of $p=\alpha+\beta$ and $q=-\alpha\beta$ i.e. $p$ and $q$ are the roots of the quadratic equation $x^2-px-q=0$.
After substituting in the first equation we get $\Delta_n =(\alpha+\beta)\Delta_{n-1} - \alpha\beta\Delta_{n-2}$
We get that there's a geometric progression in the last equation and by writing all down we get that
$$ \Delta_n=\frac{\beta^{n-1}(\Delta_2-\alpha\Delta_1)-\alpha^{n-1}(\Delta_2-\beta\Delta_1)}{\beta-\alpha} $$
Now, here's the thing I don't understand. In the book it is written that the last formula is easily remembered with the formula $$\Delta_n=c_1\alpha^n+c_2\beta^n$$ where $c_1$ and $c_2$ are taken from the same with $n=1, n=2$.
I really don't get how we got this representation using $c_1$ and $c_2$ and what exactly $c_1$ and $c_2$ actually equal.
Thanks in advance!
In the first part of your calculation you showed that $\Delta_n=c_1\alpha^n+c_2\beta^n$ for some $c_1,c_2$. How to find them? If you plug in $n=1,2$:
$$ \begin{array}{l} \Delta_1 = c_1\alpha+c_2\beta \\ \Delta_2 = c_1\alpha^2+c_2\beta^2 \\ \end{array} $$
and you can compute $c_1,c_2$ from this system of two equations.
Personally I think it is even easier to take $n=0$ and the equation $\Delta_0=c_1+c_2$.