问题 解答题
已知等差数列{an}的首项a1=1,公差d>0,且a2,a5,a14成等比数列.
(1)求数列{an}的通项公式;
(2)设bn=
1
n(an+3)
(n∈N*),Sn=b1+b2+…+bn
,求Sn
1
36
答案

(1)∵等差数列{an}的首项a1=1,公差d>0,且a2,a5,a14成等比数列,

(a1+d)(a1+13d)=(a1+4d)2

整理得:2a1d=d2

∵a1=1,解得d=2(d=0舍去)

an=2n-1(n∈N*)

(2)bn=

1
n(an+3)
=
1
2n(n+1)
=
1
2
(
1
n
-
1
n+1
),

∴Sn=b1+b2+…+bn

=

1
2
[(1-
1
2
)+(
1
2
-
1
3
)+…+(
1
n
-
1
n+1
)]

=

1
2
(1-
1
n+1
),

∴当n=1时,Sn取最小值S1=

1
2
(1-
1
2
)=
1
4
1
36

Sn

1
36

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