Kenneth Rosen Edition 7 Exercise 8.2 Question 19 (Page No. 525)

Question

Solve the recurrence relation $a_{n} = −3a_{n−1} − 3a_{n−2} − a_{n−3}\:\text{with}\: a_{0} = 5, a_{1} = −9,\:\text{and}\: a_{2} = 15.$

Answer 1

characterstick equation

$r^{3}+3r^{2}+3r+1=0$

$(r+1)^{3}=0$

$r=-1,-1,-1$

than solution

$a_n=(A+Bn+Cn^{2})(-1)^{n}$ eq->.....(1)

put n=0,1,2 then

5=A

9=A+B+C

15=A+2B+4C

after solving A=5,B=3,C=1

THEN final solution

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