Arrange the steps to solve the recurrence relation $a_n = a_{n-1} + 6a_{n-2}$ for $n \ge 2$ together with the initial conditions $a_0 = 3$ and $a_1 = 6$ in the correct order.
Rank the options below.
$a_1 = 3/5$ and $a_2 = 12/5$
Therefore, $a_n = (3/5) \cdot 2^n + (12/5) \cdot 3^n$
$3 = c_1 + c_2$
$6 = 2c_1 + 3c_2$
$a_n = c_1 \cdot 2^n + c_2 \cdot 3^n$
$r^2 - r - 6 = 0$ and $r = -2, 3$
