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Now solve for \(r\text{:}\)

so by factoring, \(r = -2\) or \(r = 3\) (or \(r = 0\text{,}\) although this does not help us). This tells us that \(a_n = (-2)^n\) is a solution to the recurrence relation, as is \(a_n = 3^n\text{.}\) Which one is correct? They both are, unless we specify initial conditions. Notice we could also have \(a_n = (-2)^n + 3^n\text{.}\) Or \(a_n = 7(-2)^n + 4\cdot 3^n\text{.}\) In fact, for any \(a\) and \(b\text{,}\) \(a_n = a(-2)^n + b 3^n\) is a solution (try plugging this into the recurrence relation). To find the values of \(a\) and \(b\text{,}\) use the initial conditions.