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 .
We can find the values of C and D by using the initial conditions a0 = 13 and a1 = -1.
a0 = C · 30 + D · (-2)0 = C + D = 13
a1 = C · 31 + D · (-2)1 = 3C - 2D = -1
Solving the two equations C + D = 13 and 3C - 2D = -1 we find that C = 5 and D = 8.
Hence the sequence a0, a1, a2,
... defined by the explicit formula an = 5 · 3n + 8· (-2)n,
satisfies the relation ak = ak-1 + 6 ak-2 and the
initial conditions a0 = 13 and a1 = -1.