Kenneth Rosen Edition 7 Exercise 8.2 Question 15 (Page No. 525)

Question

Find the solution to $a_{n} = 2a_{n−1} + 5a_{n−2} − 6a_{n−3}\: \text{with}\: a_{0} = 7, a_{1} = −4,\:\text{and}\: a_{2} = 8.$

Answer 1

characterstick equation

$r^{3}-2r^{2}-5r+6=0$

$(r+2)(r-3)(r-1)=0$

$r=-2,1,3$

then solution

$a_n=A(1)^{n}+B(3)^{n}+C(-2)^{n}$

put n=0,1,2

7=A+B+C

-4=A+3B-2C

8=A+9B+4C

after solution A=5,B=-1,C=3

$a_n=5(1)^{n}-1(3)^{n}+3(-2)^{n}$