# Solving a recurrence relation

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## Homework Statement

Solve the recurrence relation
an = 5an−1 − 3an−2 − 9an−3 for n ≥ 3
with initial values a0 = 0, a1 = 11, and a2 = 34.

its given lol

## The Attempt at a Solution

I found that the characteristic equation for this rr is x3 - 5x2 + 3x + 9 and found that the characteristic roots are 3, 3, -1...because we have 2 indistinct roots, I multiplied one of the 3 terms by n to get

an = r3n + sn3n - t
and so plugging back into the give rr we have

r3n + sn3n - t = 5(r3n-1 + s(n-1)3n-1 - t) - 3(r3n-2 + s(n-2)3n-2 - t) - 9(r3n-3 + s(n-3)3n-3 - t)

I'm thinking that in order to solve this, we're going to have to set this up as a system of equations, but I'm not sure how to do that with what I have. Any hints/tips/ suggestions on where to go next would be very helpful.

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an = r3n + sn3n - t

Given this, then r,s and t must be equal to what for you to have a0 = 0, a1 = 11, and a2 = 34? This above expression must be valid for all n, after all, not only for $n\geq 3$.

By the way, you claim that one of your roots is -1; are you sure that the above is entirely correct?

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By the way, you claim that one of your roots is -1; are you sure that the above is entirely correct?

To see if a root exists, we would plug it into the characteristic equation. When -1 is plugged into the equation, we obtain 0, therefore it is a root of the equation. Corollary, I saw that 3 was a root in same fashion, and found that it was a root of multiplicity 2 when I plugged it into the derivative. This was how we were showed to find the roots.

I don't have any doubt that -1 is a root: it is. But, if 3 is a root (forget the multiplicity for a moment) and it gives rise to a term 3n in the solution, then what would be the term corresponding to -1?