(Ⅰ)由an+2=3an+1-2an⇒an+2-an+1=2(an+1-an),a1=1,a2=3,从而可证数列{an+1-an}是等比数列;
(Ⅱ)由(Ⅰ)得an+1-an=2n(n∈N*),利用累加法,借助等比数列的求和公式即可求得数列{an}的通项公式.
证明:(Ⅰ)∵an+2=3an+1-2an,
∴an+2-an+1=2(an+1-an),
∴=2(n∈N*).
∵a1=1,a2=3,
∴数列{an+1-an}是以a2-a1=2为首项,2为公比的等比数列.
(Ⅱ)由(Ⅰ)得an+1-an=2n(n∈N*),
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=2n-1+2n-2+…+2+1
=2n-1(n∈N*).