a(n+1)-5an+6a(n-1)=0
a(n+1)-2an = 3(an-2a(n-1))
[a(n+1)-2an]/(an-2a(n-1)) =3
[a(n+1)-2an]/(a2-2a1) =3^(n-1)
a(n+1)-2an = -3^(n-1)
an-2a(n-1) =-3^(n-2)
an + (1/3).3^n = 2[ a(n-1) + (1/3).3^(n-1)]
[an + (1/3).3^n]/[ a(n-1) + (1/3).3^(n-1)] =2
[an + (1/3).3^n]/[ a2 + (1/3).3^2] = 2^(n-2)
an + (1/3).3^n = 2^n
an = 2^n-3^(n-1)求出的特征根是2和3？？這個題目能用特征根方程做嗎？滿足A(n+2)=pA(n+1)-qAn形式嗎不
a(n+1)-5an+6a(n-1)=0

The aux. equation
x^2-5x+6 =0
(x-2)(x-3)=0
x= 2 or 3
特征根是2和3


an = A(3)^n +B(2)^n
a1 = 3A+2B = 1(1)
a2 = 9A+ 4B =1(2)
solving
A = -1/3 , B= 1
an = -3^(n-1) +2^n