$\newcommand{\nfrac}[2]{\frac{\displaystyle{#1}}{\displaystyle{#2}}}$
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 $F(n) = (-2)^n + 3 (5)^n$.