已知数列〔an〕中，a1=1,a2=2，a(n+1)=2an+a(n-1)（n属于N+），求数列的通项公式

a(n+1) + xa(n) = (2+x)[a(n) + xa(n-1)],
1 = (2+x)x, 0 = x^2 + 2x - 1 = (x+1)^2 - 2 = (x+1+2^(1/2))(x+1-2^(1/2)),
x = 2^(1/2)-1或x = -1-2^(1/2).

b(n) = a(n+1) + [2^(1/2) - 1]a(n),
b(n+1) = [2^(1/2)+1]b(n),
{b(n)}是首项为b(1)=a(2)+[2^(1/2)-1]a(1) = 2^(1/2) + 1,公比为[2^(1/2)+1]的等比数列。
b(n) = [2^(1/2)+1][2^(1/2)+1]^(n-1) = [2^(1/2)+1]^n,

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

c(n) = a(n)[2^(1/2)-1]^(n-1),
1 = c(n+1) + [2^(1/2)-1]^2c(n),
c(n+1) + y = -[2^(1/2)-1]^2[c(n) + y],
1 = -y - [2^(1/2)-1]^2y, y = -1/[1 + [2^(1/2)-1]^2] = -1/[1 + 3 - 2*2^(1/2)] = -1/[4 - 2*2^(1/2)] = -[2+2^(1/2)]/4,

c(n+1) - [2+2^(1/2)]/4 = -[3-2*2^(1/2)][c(n) - [2+2^(1/2)]/4],
{c(n)-[2+2^(1/2)]/4}是首项为c(1)-[2+2^(1/2)]/4 = a(1)-[2+2^(1/2)]/4 = 1 - [2+2^(1/2)]/4 = [2-2^(1/2)]/4,公比为[2*2^(1/2) -3]的等比数列。
c(n)-[2+2^(1/2)]/4 = [2-2^(1/2)][2*2^(1/2)-3]^(n-1)