(Ⅰ)∵an+1=f(an),∴an=f-1(an+1)
∵任意实数x都有f(x)+
2
3f−1(x)=
5
3x,∴f(an+1)+
2
3f−1(an+1)=
5
3an+1
∵an+1=f(an),即an=f-1(an+1),∴f(an+1)=an+2,f-1(an+1)=an
∴an+2+
2
3an=
5
3an+1,即an+2−an+1=
2
3(an+1−an)
∵bn=an+1-an(n∈N*),a1=1,a2=
5
3]
∴数列{bn}是以a2−a1=
5
3−1=
2
3为首项,以q=
2
3为公比的等比数列
故数列{bn}的通项为bn=(
2
3)n
(Ⅱ)(文)由bn=an+1−an=(
2
3)n得an-a1=(an-an-1)+(an-1-an-2)+…+(a2-a1)
=(
2
3)n−1+(
2
3)n−2+…+(
2
3)2+
2
3=2[1−(
2
3)n−1]
又∵a1=1,∴an=3−
2n
3n−1(n∈N*),即数列{an}是递增数列,且an<3(n∈N*)
∴满足
m−1
2m+1>an对所有n∈N*恒成立的参数m必须满足