已知数列{an}中,a1=1,an=an-1?3n-1(n≥2,n∈N*),数列{bn}的前n项和Sn=log3(an9n)(n∈N*).(I)求

2025-05-13 21:38:30
推荐回答(1个)
回答(1):

(I)∵log3an=log3an-1?3n-1,两边取以3为底的对数得log3an=log3an-1+(n-1)移向得log3an-log3an-1=n-1,
log3a2-log3a1=1,
log3a3-log3a2=2,

log3an-log3an-1=n-1,
以上各式相加得(n≥2)
log3an?log3a1=1+2+…+(n?1)=

n(n?1)
2
,log3an
n(n?1)
2
,且对n=1时也成立.
Sn=log3(
an
9n
)=
n2?5n
2
(n∈N*)

∴b1=S1=-2,
当n≥2时,bn=Sn-Sn-1=n-3,且对n=1时也成立
∴数列{bn}的通项公式bn=n-3(n∈N*).
(II)设数列{|bn|}的前n项和为Tnbn=n?30≤0即n≤3时,Tn=?(b1+b2+…+bn)=?S  n
5n?n2
2
n>3时,Tn=?(a1+a2+a3)+(a4+a5+…+an)=Sn?2S3
n2?5n+12
2