a(n+1)-3an+2a(n-1)=0
a(n+1)-an=2[an-a(n-1)]
[a(n+1)-an]/[an-a(n-1)]=2
{an-a(n-1)}是公比为2的等比数列
an-a(n-1)=(a2-a1)*2^(n-2)=(2-1)*2^(n-2)= 2^(n-2)
a(n-1)-a(n-2)=2^(n-3)
...
a2-a1=2^(0)
两边相加得:
an-a1=2^(n-2)+2^(n-3)+...+2^(0)= (1-2^(n-1))/(1-2)=2^(n-1)-1
an=a1+2^(n-1)-1=2^(n-1)
an=2^(n-1).