a(n+1)-an=2[an-a(n-1)]
bn=2b(n-1),b1=a2-a1=3,bn是首项为3,公比为2的等比数列
bn=3*2^(n-1)
a(n+1)-an=bn=3*2^(n-1)
a(n+1)=3*2^(n-1)+an
=3*2^(n-1)+3*2^(n-2)+a(n-1)
=3*2^(n-1)+3*2^(n-2)+3*2^(n-3)+a(n-2)
=...
=3*2^(n-1)+3*2^(n-2)+3*2^(n-3)+...+3*2^0+a1
=1+3*[1+2^1+2^2+...+2^(n-1)]
=1+3*(2^n-1)
an=1+3*[2^(n-1)-1]