4. Consider the second-order linear homogeneous recurrence relation bk = 5 bk-1 - 4 bk-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 - 5 · t + 4 = 0.
By factoring, t2 - 5 · t + 4 = (t - 4) (t - 1), so the two values of t which satisfy the characteristic equation are t = 4 and t = 1.
Hence the two sequences which satisfy the recurrence relation bk = 5 bk-1 - 4 bk-2 are 1, 4, 16, 64, ... 4n, ... and 1, 1, 1, ..., 1n, ... .
By Lemma 8.3.2, any sequence b0, b1, b2, ... that satisfies an explicit formula of the form bn = C · 4n + D · 1n, where C and D are real numbers, also satisfies the relation bk = 5 bk-1 - 4 bk-2 .
We can find the values of C and D by using the initial conditions b0 = 2 and b1 = 3.
b0 = C · 40 + D · 10 = C + D = 2
b1 = C · 41 + D · 11 = 4C + D = 3
Solving the two equations C + D = 2 and 4C + D = 3 we find that C = 1/3 and D = 5/3.
Hence the sequence b0, b1, b2,
... defined by the explicit formula bn = 1/3 · 4n
+ 5/3 · 1n,
satisfies the relation bk = 5 bk-1 - 4 bk-2 and the
initial conditions b0 = 2 and b1 = 3.