题目内容
等差数列{an}是递增数列,前n项和为Sn,且a1,a3,a9成等比数列,S5=a52.(1)求数列{an}的通项公式;
(2)若数列{bn}满足bn=
,求数列{bn}的前99项的和.
解:(1)设数列{an}公差为d(d>0),
∵a1,a3,a9成等比数列,∴a32=a1a9.
(a1+2d)2=a1(a1+8d),d2=a1d.
∵d≠0,∴a1=d. ①
∵S5=a52,∴5a1+
·d=(a1+4d)2. ②
由①②得a1=
,d=
.
∴an=
+(n-1)×
=
n.
(2)bn=
,
∴b1+b2+b3+…+b99
=
[99+(1-
)+(
-
)+…+ (
-
)]
=
(100-
)=
.
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