Kenneth Rosen Edition 7 Exercise 8.2 Question 13 (Page No. 525)

Question

Find the solution to $a_{n} = 7a_{n−2} + 6a_{n−3}\:\text{with}\: a_{0} = 9, a_{1} = 10, \text{and}\: a_{2} = 32.$

Answer 1

$r^{ 3 }− 7r − 6 = 0$

 $(r + 1)(r 2 − r − 6) = 0$.

Continue factoring and we get $(r + 1)(r − 3)(r + 2) = 0$.

We know our characteristic roots: r1 = −1, r2 = 3, r3 = −2.

Our general solution using Theorem 3 is: an = $α1(−1)^{n} + α2(3)^{n } + α3(−2)^{n}$ .

Find α1, α2, α3 by using the initial conditions.

For a0 = 9

$9 = α1(−1)^0 + α2(3)^ 0 + α3(−2)^0$

9 = α1 + α2 + α3

For a1 = 10

$10 = α1(−1)^{1} + α2(3 )^{1} + α3(−2)^{1}$

10 = −α1 + 3α2 − 2α3

For a2 = 32

$32 = α1(−1)^{2} + α2(3)^{2} + α3(−2)^{2}$

32 = α1 + 9α2 + 4α3

Solve the system of equations:

9 = α1 + α2 + α3

10 = −α1 + 3α2 − 2α3

32 = α1 + 9α2 + 4α3

Our solution is $an = 8(−1)^{n} + 4(3)^{n }+ (−3)(−2)^{n}$