若数列{an}是首项为19,公差为-2的等差数列(1)求数列{an}的通项公式及前n项和Sn(2)设{bn-an}是以1为

2025-05-14 15:56:05
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回答1:

(1)∵数列{an}是首项为19,公差为-2的等差数列,
∴an=19+(-2)(n-1)=-2n+21.
Sn=

n(19+21?2n)
2
=-n2+20n.
(2)∵{bn-an}是以1为首项,以3为公比的等比数列,
∴bn-an=3n-1,∴bnan+3n?1=-2n+21+3n-1
∴Tn=Sn+
3n?1
3?1
=?n2+2n+
3n?2
2