2Sn=3an-3
2S(n-1)=3a(n-1)-3
相减得2an=3an-3a(n-1)
an=3a(n-1)
2S1=2a1=3a1-3
a1=3
所以{an}是以3为首项,3为公比的等比数列
an=3×3^(n-1)=3^n
bn=1/[log3an·log3a(n+1)]
=1/[log33^n·log33^(n+1)]
=1/[n(n+1)]
=1/n-1/(n+1)
Tn=1-1/2+1/2-1/3+…+1/n-1/(n+1)
=1-1/(1+n)
=n/(1+n)
2Sn=3an-3
2S(n-1)=3a(n-1)-3
相减得2an=3an-3a(n-1)
an=3a(n-1)
2S1=2a1=3a1-3
a1=3
所以{an}是以3为首项,3为公比的等比数列
an=3×3^(n-1)=3^n
bn=1/[log3an·log3a(n+1)]
=1/[log33^n·log33^(n+1)]
=1/[n(n+1)]
=1/n-1/(n+1)
Tn=1-1/2+1/2-1/3+…+1/n-1/(n+1)
=1-1/(1+n)
=n/(1+n)