a(n+1)=sn/2
sn=2a(n+1)
s(n-1)=2an
sn-a(n-1)=2a(n+1)-2an
an=2a(n+1)-2an
2a(n+1)=3an
a(n+1)/an=3/2
所以an是以3/2为公比的等比数列
an=a1q^(n-1)
=(3/2)^(n-1)
a(n+1)=(3/2)^n
bn=log3/2[a(n+1)]
=log3/2[(3/2)^n]
=n
b1=1
b(n+1)=n+1
Tn=1/b1b2+1/b2b3+.+1/bnb(n+1)
=1/1*2+1/2*3+.+1/n(n+1)
=1-1/2+1/2-1/3+.+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)