由题意得:Sn=a(Sn-an+1)(a>0)
则易得:Sn=[a/(a-1)]*(an-1)
根据:an=Sn-S(n-1)(n>1)
所以:an=[a/(a-1)]*(an-1)-[a/(a-1)]*[a(n-1)-1]=[a/(a-1)]*[an-a(n-1)]
得:an/a(n-1)=a
因此:数列{an}是以a为公比的等比数列
an=a1*a^(n-1)
又因为:S1=a(S1-a1+1)且S1=a1
即:a1=a
所以:an=a^n
4a3是a1与2a的等差中项,则有8a3=a1+2a
即有8a^3=a+2a=3a
a^2=3/8
a=根号6/4
故an=(根号6/4)^n
bn=2n+1/an=2n+(4/根号6)^n
Tn=2(1+n)n/2+4/根号6*((4/根号6)^n-1)/(4/根号6-1)
=n(n+1)+4((4/根号6)^n-1)/(4-根号6)