5. Consider the second-order linear homogeneous recurrence relation ak = ak-1 + 6 ak-2.
To find a sequence which satisfies this recurrence relation and also satisfies the initial conditions, you need to start by finding the two sequences which satisfy the relation by solving the characteristic equation. The characteristic equation of the relation is t2 - t - 6 = 0.
By factoring, t2 - t - 6 = (t - 3) (t + 2), so the two values of t which satisfy the characteristic equation are t = 3 and t = -2.
Hence the two sequences which satisfy the recurrence relation ak = ak-1 + 6 ak-2 are 1, 3, 9, 27, ... 3n, ... and 1, -2, 4, ..., (-2)n, ... .
By Lemma 8.3.2, any sequence a0, a1, a2, ... that satisfies an explicit formula of the form an = C · 3n + D · (-2)n, where C and D are real numbers, also satisfies the relation ak = ak-1 + 6 ak-2 .
Now find the values of C and D by using the initial conditions a0 = 13 and a1 = -1.