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a)

The characteristic polynomial of this recurrence relation is

$r^3-6r^2+11r-6=0$

$r^2(r-1)-5r(r-1)+6(r-1)=0$

$(r-1)(r^2-5r+6)=0$

$(r-1)(r-2)(r-3)=0$

$r_1=1, r_2=2, r_3=3$

Hence, the solutions to this recurrence relation are of the form

$a_n=\alpha_1\cdot1^n+\alpha_2\cdot2^n+\alpha_3\cdot3^n$

Use the initial conditions

$a_0=4=\alpha_1+\alpha_2+\alpha_3$$a_1=6=\alpha_1+2\alpha_2+3\alpha_3$$a_2=12=\alpha_1+4\alpha_2+9\alpha_3$

$\alpha_1+\alpha_2+\alpha_3=4$$\alpha_2+2\alpha_3=2$$2\alpha_2+6\alpha_3=6$

$\alpha_1+\alpha_2+\alpha_3=4$$\alpha_2+2\alpha_3=2$$\alpha_2+3\alpha_3=3$

$\alpha_1=3$$\alpha_2=0$$\alpha_3=1$

Hence, the unique solution to this recurrence relation and the given initial conditions is the sequence $\{a_n\}$ with

$a_n=3+3^n$

b)

The characteristic polynomial of this recurrence relation is

$r^3-r^2-9r+9=0$

$r^2(r-1)-9(r-1)=0$

$(r-1)(r-3)(r+3)=0$

$r_1=1, r_2=-3, r_3=3$

Hence, the solutions to this recurrence relation are of the form

$a_n=\alpha_1\cdot1^n+\alpha_2\cdot(-3)^n+\alpha_3\cdot3^n$

Use the initial conditions

$a_0=6=\alpha_1+\alpha_2+\alpha_3$$a_1=0=\alpha_1-3\alpha_2+3\alpha_3$$a_2=30=\alpha_1+9\alpha_2+9\alpha_3$

$\alpha_1+\alpha_2+\alpha_3=6$$4\alpha_2-2\alpha_3=6$$8\alpha_2+8\alpha_3=24$

$\alpha_1+\alpha_2+\alpha_3=6$$2\alpha_2-\alpha_3=3$$\alpha_2+\alpha_3=3$

$\alpha_1=3$$\alpha_2=2$$\alpha_3=1$

Hence, the unique solution to this recurrence relation and the given initial conditions is the sequence $\{a_n\}$ with

$a_n=3+2\cdot(-3)^n+3^n$