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I have a recurrence relation: $A_{n} = -2A_{n-1} + 15A_{n-2}$ , with $a_{1} = 10$ and $a_{2} = 70$.

This would be a linear homogenous recurrence relation of degree 2, right?

Using the relation I found the first five terms to be $10, 70, 10, 1030$ and $1910$ respectively.

I then tried to solve the relation and got $x^{2} + 2x -15 = 0$ as my characteristic equation, and got roots of $5$ and $3$.

So, $A_{n} = a_{5}^{n} + b_{3}^{n}$ , right?

I solved this $2 \times 2$ system:

$a_{1} = a_{5}^{2} + b_{3} = 10$

$a_{2} = a_{5}^{2} + b_{3}^{2} = 70$

and got: $a = 4, b = -\frac{10}{3}$

So my explicit equation came out to $A_{n} = 4(5)^n - \frac{10}{3} (3)^{n}$

When i use the equation for $a_{1}$ and $a_{2}$ the answers match those that I get using the recurrence relation, but $a_{3}$ and onward do not give the same results.

for example, for $a_{5}$ I got 1910 from the recurrence relation and 11690 from my formula.

I must have messed up somewhere, but I can't figure out where.

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  • $\begingroup$ I think the edit is wrong $\endgroup$ – Mike Feb 1 '15 at 20:21
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The roots of the characteristic equations $$x^2+2x-15=0$$ are $3$ and $-5$, so you have to use $(-5)^n$ instead of $5^n$

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  • $\begingroup$ Thanks Peter, I was stuck in Einstellung for a bit there. Got it sorted out thanks to your answer. $\endgroup$ – Mike Feb 1 '15 at 20:50

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