问题
解答题
已知等比数列{an}中,a1=
(I)Sn为{an}的前n项和,证明:Sn=
(II)设bn=log3a1+log3a2+…+log3an,求数列{bn}的通项公式. |
答案
证明:(I)∵数列{an}为等比数列,a1=
,q=1 3 1 3
∴an=
×(1 3
)n-1=1 3
,1 3n
Sn=
=
(1- 1 3
)1 3n 1- 1 3 1- 1 3n 2
又∵
=1-an 2
=Sn1- 1 3n 2
∴Sn=1-an 2
(II)∵an=1 3n
∴bn=log3a1+log3a2+…+log3an=-log33+(-2log33)+…-nlog33
=-(1+2+…+n)
=-n(n+1) 2
∴数列{bn}的通项公式为:bn=-n(n+1) 2