Chapter 04 - Section 02

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Exercises

9

We follow the standard recipe. $$ \begin{align*} r^2 &= 3r + 10 \\ r^2 - 3r - 10 &= 0 \\ (r + 2)(r - 5) &= 0 \\ \end{align*} $$ $r = -2$ or $r = 5$. The general solution form is $\alpha_1(-2)^n + \alpha_2(5)^n$. From the base cases: $$ \begin{align*} F(0) &= 4 \\ &= \alpha_1 + \alpha_2 \\ F(1) &= 13 \\ &= -2 \alpha_1 + 5 \alpha_2 \\ \end{align*} $$ Which by substitution implies $\alpha_1 = 1$ and $\alpha_2 = 3$. Therefore $F(n) = (-2)^n + 3 (5)^n$.