Thread: Recurrence Relation - Next Step

1. Recurrence Relation - Next Step

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.

2. You have the quadratic we need to work with OK.

Since the roots are -2 and 3, we have

$A\cdot (-2)^{n}+B\cdot 3^{n}$

Now, use the initial conditions, $a_{0}=3, \;\ a_{1}=6$

If n=0, we have $A\cdot (-2)^{0}+B\cdot 3^{0}=A+B=3$

If n=1, we have $A\cdot (-2)^{1}+B\cdot 3^{1}=-2A+3B=6$

The two equations to solve are:

$A+B=3$
$-2A+3B=6$

$A=\frac{3}{5}, \;\ B=\frac{12}{5}$

$\boxed{\frac{3}{5}\cdot (-2)^{n}+\frac{12}{5}\cdot 3^{n}}$