Recurrence Relation

**vexiked** · March 22nd 2009, 09:03 AM

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.

**galactus** · March 22nd 2009, 09:41 AM

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:

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

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

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

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