Kenneth Rosen Edition 7 Exercise 8.2 Question 12 (Page No. 525)

Question

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

Answer 1

characterstick equation

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

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

$r=1,-1,2$

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

put vakue in equation

3=A+B+C

6=A-B+2C

0=A+B+4C

THEN VALUE OF A=6,B=-2,C=-1

$a_n=6(1)^{n}-2(-1)^{n}-(2)^{n}$