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=
(n∈N*),Tn=b1+b2+…+bn,求Tn. 1 n(3−lgan)
答
(1)由题知2lga2=lga1+(1+lga4),即:lga22=lg10a1a4,
则a22=10a1a4=10a12q3,
∵a1>0,q2>0,
∴q=
.(3分)1 10
又a1a2a3=1,
∴a13q3=a13(
)3=1,1 10
∴a13=1000,
∴a1=10,(6分)
∴an=10×(
)n−1=102-n,(8分)1 10
(2)bn=
=1 n(3−lgan)
=1 n(n+1)
-1 n
(10分)1 n+1
∴Tn=b1+b2+…+bn
=(1-
)+(1 2
-1 2
)+…+(1 3
-1 n
)1 n+1
=1-
1 n+1
=
(12分)n n+1