...均为正数,前n项和为Sn,且Sn=an(an+1)2(n∈N*).(1)求数列{an}的通项...
发布网友
发布时间:2024-10-05 15:16
我来回答
共1个回答
热心网友
时间:2024-10-05 15:14
(1)Sn=an(an+1)2,n∈N+,当n=1时,S1=a1(a1+1)2,∴a1=1…(1分)
∵2Sn=an2+an,
当n≥2时,2Sn-1=an?12+an-1,
两式相减得:2an=2(Sn?Sn?1)=a2n?a2n?1+an?an?1,…(3分)
∴(an+an-1)(an-an-1-1)=0,
∵an+an-1>0,
∴an-an-1=1,n≥2,…(5分)
∴数列{an}是等差数列,∴an=n…(6分)
(2)由(1)Sn=n(n+1)2,
∴bn=?2Sn(n+1)?2n=?n2n,…(8分)
∴?Tn=12+222+…+n?12n?1+n2n,…(9分)
?2Tn=1+22+…+n?12n?2+n2n?1,…(10分)
∴Tn=?1?12?…?12n?1+n2n
=?1?12n1?12+n2n
=?2+12n?1+n2n=?2+n+22n.…(12分).