若数列{an}是等比数列,an>0,公比q≠1,已知lga2是lga1和1+lga4的等差中项,且a1a2a3=1. (1)求{an}的通项公式; (2)设bn=1/n(3−lgan)(n∈N*),Tn=b1+b2+…+bn,求Tn.

问题描述:

若数列{an}是等比数列,an>0,公比q≠1,已知lga2是lga1和1+lga4的等差中项,且a1a2a3=1.
(1)求{an}的通项公式;
(2)设bn=

1
n(3−lgan)
(n∈N*),Tn=b1+b2+…+bn,求Tn

(1)由题知2lga2=lga1+(1+lga4),即:lga22=lg10a1a4
a22=10a1a4=10a12q3
∵a1>0,q2>0,
∴q=

1
10
.(3分)
又a1a2a3=1,
a13q3=a13(
1
10
)
3
=1,
a13=1000,
∴a1=10,(6分)
∴an=10×(
1
10
)
n−1
=102-n,(8分)
(2)bn=
1
n(3−lgan)
=
1
n(n+1)
=
1
n
-
1
n+1
(10分)
∴Tn=b1+b2+…+bn
=(1-
1
2
)+(
1
2
-
1
3
)+…+(
1
n
-
1
n+1

=1-
1
n+1

=
n
n+1
(12分)