Answer
$a_n = −5 . 2^n + n . 2^{n−1} + 13 . n^2
. 2^{n−1}$
Work Step by Step
The characteristic equation of this recurrence relation is $$r^3−6r^2+12r−8 $$
Because $r^3−6r^2+12r−8 = (r−2)^3$ so $(r−2)^3=0$.
We know from Theorem 4 on page 519 that
$$a_n = γ_12^n + γ_2n2^n + γ_3n^22^n$$
for some constants $γ_1, γ_2, γ_3$.
By substituting $n = 0, 1, 2$ into this equation, we obtain the linear system
$$γ_1 = −5$$
$$2γ_1 + 2γ_2 + 2γ_3 = 4$$
$$4γ_1 + 8γ_2 + 16γ_3 = 88$$
The solutions are $γ_1 = −5 , γ_2 = \frac{1}{2}, γ_3 = \frac{13}{2}$, and therefore
$$a_n = −5 . 2^n + n . 2^{n−1} + 13 . n^2
. 2^{n−1}$$
for $n = 0, 1, 2, 3, . . . $
