问题 解答题
已知公差不为0的等差数列{an}的首项为4,设数列的前n项和为Sn,且
1
a1
1
a2
1
a4
成等比数列.
(1)求数列{an}的通项公式an及Sn
(2)记An=
1
S1
+
1
S2
+
1
S3
+…+
1
Sn
,Bn=
1
a1
+
1
a2
+
1
a22
+…+
1
a2n-1
,当n≥2时,试比较An与Bn的大小.
答案

(1)设等差数列{an}的公差为d,由(

1
a2
)2=
1
a1
1
a4

得(a1+d)2=a1(a1+3d),因为d≠0,所以d=a1=4,

所以an=4n,Sn=4n+

4n(n-1)
2
=2n(n+1);

(2)∵

1
Sn
=
1
2
(
1
n
-
1
n+1
)

∴An=

1
S1
+
1
S2
+…+
1
Sn
=
1
2
(1-
1
2
+
1
2
-
1
3
+…+
1
n
-
1
n+1
)=
1
2
(1-
1
n+1
)

a2n-1=4•2n-1=2n+1

Bn=

1
a1
+
1
a2
+…+
1
a2n-1
=
1
4
1-(
1
2
)n
1-
1
2
=
1
2
(1-
1
2n
)

当n≥2时,2n=

C0n
+
C1n
+…+
Cnn
>n+1,

1-

1
n+1
<1-
1
2n

所以An<Bn

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