The question is: An = An-1 + 6An-2 for n>=2, a0=3, a1=6
So far I have...
r^2-r-6=0 so the roots are r=-2,3.
I am stuck here and not sure where to go. Can someone please help me! Thanks in advance.
You have the quadratic we need to work with OK.
Since the roots are -2 and 3, we have
$\displaystyle A\cdot (-2)^{n}+B\cdot 3^{n}$
Now, use the initial conditions, $\displaystyle a_{0}=3, \;\ a_{1}=6$
If n=0, we have $\displaystyle A\cdot (-2)^{0}+B\cdot 3^{0}=A+B=3$
If n=1, we have $\displaystyle A\cdot (-2)^{1}+B\cdot 3^{1}=-2A+3B=6$
The two equations to solve are:
$\displaystyle A+B=3$
$\displaystyle -2A+3B=6$
$\displaystyle A=\frac{3}{5}, \;\ B=\frac{12}{5}$
$\displaystyle \boxed{\frac{3}{5}\cdot (-2)^{n}+\frac{12}{5}\cdot 3^{n}}$