记公差不为0的等差数列{an}的前n项和为Sn,S3=9,a3,a5,a8成等比数列.(Ⅰ) 求数列{an}的通项公式an

2025-04-06 13:21:54
推荐回答(1个)
回答1:

(Ⅰ) 由已知得:S3=9,a52a3?a8

3a1+
3×2
2
d=9
(a1+4d)2=(a1+2d)?(a1+7d)
,解得:a1=2,d=1.
∴an=n+1,
Sn
n(2+n+1)
2
n2
2
+
3
2
n

(Ⅱ)由题知cn=2n(
2
n+1
?λ)

若使{cn}为单调递减数列,
则cn+1-cn=2n+1(
2
n+2
?λ)
-2n(
2
n+1
?λ)
=2n(
4
n+2
?
2
n+1
?λ)<0
对一切n∈N*恒成立,
即:
4
n+2
?
2
n+1
?λ<0?λ>(
4
n+2
?
2
n+1
)max

4
n+2
?
2
n+1
=
2n
(n+2)(n+1)
2n
n2+3n+2
2
n+
2
n
+3

当n=1或2时,(
4
n+2
?
2
n+1
)max
=
1
3

λ>
1
3

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