1.
a2=(1/2)S1=(1/2)a1=(1/2)×1=1/2
a(n+1)=(1/2)Sn
Sn=2a(n+1)
n≥2时,an=Sn-S(n-1)
=2a(n+1)-2an
2a(n+1)=3an
a(n+1)/an=3/2,定值.n≥2时,数列是以1/2为首项,3/2为公比的等比数列.
an=(1/2)(3/2)^(n-2)=3^(n-2) /2^(n-1)
n=1时,a1=(1/2)×(3/2)^(1-2)=(1/2)(2/3)=1/3≠1
数列{an}的通项公式为
an=1 n=1
3^(n-2) /2^(n-1) n≥2
2.
bn=log(3/2)[3a(n+1)]=log(3/2)[3×3^(n-1) /2^n]=log(3/2)[(3/2)^n]=n
1/[bnb(n+1)]=1/[n(n+1)]=1/n -1/(n+1)
Tn=1/(b1b2)+1/(b2b3)+...+1/[bnb(n+1)]
=1/1-1/2+1/2-1/3+...+1/n -1/(n+1)
=1- 1/(n+1)
=n/(n+1)
本题的难点在于第1问,其实第1问求出来了,第2问就迎刃而解了.