Question

数列{an}:3an+2-5an+1+2an=0(n≥0,n∈N),a1=a,a2=b,求数列{an}的通项公式?

Answer 1

3a(n+2)-5a(n+1)+2a(n)=0,

3a(n+2)-3a(n+1)=2a(n+1)-2a(n),

a(n+2)-a(n+1)=(2/3)[a(n+1)-a(n)]

{a(n+1)-a(n)}是首项为a(2)-a(1)=b-a,公比为2/3的等比数列.

a(n+1)-a(n)=(b-a)(2/3)^(n-1),

(3/2)^na(n+1) - (3/2)^(n-1)*3a(n)/2 = 3(b-a)/2,

c(n)=(3/2)^(n-1)a(n),

c(n+1) - (3/2)c(n) = 3(b-a)/2,

c(n+1)=(3/2)c(n) + 3(b-a)/2,

c(n+1)+3(b-a) = (3/2)c(n) + 9(b-a)/2 = (3/2)[c(n)+3(b-a)],

{c(n)+3(b-a)}是首项为c(1)+3(b-a)=a(1)+3(b-a)=a+3(b-a)=3b-2a,公比为3/2的等比数列.

c(n)+3(b-a)=(3b-2a)(3/2)^(n-1)

c(n)=(3b-2a)(3/2)^(n-1) - 3(b-a) = (3/2)^(n-1)a(n),

a(n) = 3b-2a - 3(b-a)(2/3)^(n-1)