(Ⅰ)由Sn+1=4an+2得,当n≥2时,有Sn=4an-1+2,两式相减得出an+1=4an-4an-1,移向an+1-2an=2(an-2an-1),可证{bn}是等比数列;
(Ⅱ)由(Ⅰ)得,bn=3•2n-1,an+1-2an=3•2n-1,两边同除以2n,构造出,数列{}是首项,公差为的等差数列.通过数列{}的通项求出{an}的通项公式.
【解析】
(Ⅰ)∵a1=1,Sn+1=4an+2,
∴S2=4a1+2=a1+a2,a2=5,
∴b1=a2-2a1.=3,
另外,由Sn+1=4an+2得,当n≥2时,有Sn=4an-1+2,
∴Sn+1-Sn=(4an+2)-(4an-1+2),
即an+1=4an-4an-1,an+1-2an=2(an-2an-1),n≥2
又∵bn=an+1-2an.∴bn=2bn-1.
∴数列{bn}是首项为3,公比为2的等比数列
(Ⅱ)由(Ⅰ)得,bn=3•2n-1,
an+1-2an=3•2n-1,
∴-=,数列{}是首项,公差为的等差数列.
=+(n-1)×=n-
an=(3n-1)•2n-2