1
∵an+1=1/2Sn
∴n≥2時(shí),
an=1/2*S(n-1)
兩式相減：
a(n+1)-an=1/2*Sn-1/2*S(n-1)=1/2*an
∴a(n+1)=3/2*an
∴a(n+1)/an=3/2
∴n≥2時(shí),{an)為等比數(shù)列,公比為3/2
首項(xiàng)a2=1/2*S1=1/2
∴{an}的通項(xiàng)公式為分段公式：
an={1 (n=1)
{1/2*(3/2)^(n-2) (n≥2)
2
bn=log(1.5)[3a(n+1)]=log(3/2)[3*1/2*(3/2)^(n-1)]
=log(3/2)[(3/2)^n]=n
∴1/bnbn+1=1/[n(n+1)]=1/n-1/(n+1)
∴Tn=1/b1*b2+1/b2*b3+.+1/bnb(n+1)
=1-1/2+1/2-1/3+.+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)