(文科)已知{an}是单调递增的等差数列,首项a1=3,前n项和为Sn,数列{bn}是等比数列,首项b1=1,且a2b2=

2025-05-13 02:35:31
推荐回答(1个)
回答1:

(Ⅰ)设公差为d,公比为q,
则a2b2=(3+d)q=12①
S3+b2=3a2+b2=3(3+d)+q=20②
联立①②可得,(3d+7)(d-3)=0
∵{an}是单调递增的等差数列,d>0.
则d=3,q=2,
∴an=3+(n-1)×3=3n,bn=2n-1…(6分)
(Ⅱ)bn=2n-1,cn=n?2n-1
∴Tn=c1+c2+…+cn
Tn=1?20+2?21+3?22+…+n?2n-1
2Tn=1?21+2?22+…+(n-1)?2n-1+n?2n…(9分)
两式相减可得,-Tn=1?20+1?21+1?22+…+1?2n-1-n?2n
∴-Tn=

1?2n
1?2
?n?2n=2n-1-n?2n
∴Tn=(n-1)?2n+1…(13分)