设数列{an}的前n项积为Tn,Tn=1-an, (1)证明{1/Tn}是等差数列; (2)求数列{anTn}的前n项和Sn.
问题描述:
设数列{an}的前n项积为Tn,Tn=1-an,
(1)证明{
}是等差数列;1 Tn
(2)求数列{
}的前n项和Sn. an Tn
答
(1)由题意得Tn=1-an,①Tn+1=1-an+1,②∴由②÷①得an+1=1−an+11−an,∴an+1=12−an,∴1Tn+1-1Tn=11−an+1-11−an=11−12−an-11−an=1,又由T1=1-a1得a1=12,∴1T1=2,∴{1Tn}是首项为2,公差为1的等差数列;...