This is just a random example, I just wonder how to solve linear homogeneous recursion relations.
If we say that $b_n = b_{n−1} + 2b_{n−2}$ and $b_1 = 1$ and $b_2 = 2$, how do I find the equation for this linear homogeneous recursion relation of second degree, and find its roots?
Here is a hint for you to further move on with solving your recursion relation.
Let us say that we have a relation of the form $pa_{n} = qa_{n-1} + ra_{n-2}$. We have to find the $n^{th}$ term of the sequence $a$. For such relations, we write a characteristic equation of the form $px^2 - qx - r = 0$. For this equation we have the roots $\alpha$ and $\beta$. Hence,
$p\alpha^2 - q\alpha - r = 0$ and $p\beta^2 - q\beta - r = 0$. Multiply the first equation by $k\alpha^{n-2}$ and the second by $m\beta^{n-2}$ (where $k$ and $m$ are some constants). From doing this, you would get the two equations,
$p(k\alpha^n) - q(k\alpha^{n-1}) - r(k\alpha^{n-2}) = 0$
$p(m\beta^n) - q(m\beta^{n-1}) - r(m\beta^{n-2}) = 0$
By adding the 2 equations, you would see that $a_n = k\alpha^n + m\beta^n$ will satisfy the recursion.
Now by using the initial conditions, one can find the values $k$ and $m$. I leave the rest for you to solve.