问题 解答题
设数列{an}满足an≠0,a1=1,an=(1-2n)anan-1+an-1(n≥2),数列{an}的前n项和为Sn
(1)求数列{an}的通项公式;
(2)求证:当n≥2时,
n
n+1
Sn<2

(3)试探究:当n≥2时,是否有
6n
(n+1)(2n+1)
Sn
5
3
?说明理由.
答案

(1)∵an≠0

∴anan-1≠0(n≥2)

an
anan-1
=
(1-2n)anan-1
anan-1
+
an-1
anan-1

1
an-1
=(1-2n)+
1
an
即有
1
an
-
1
an-1
=2n-1

1
an
=
1
a1
+(
1
a2
-
1
a1
)+(
1
a3
-
1
a2
)+…+(
1
an
-
1
an-1
)=1+3+5+7+…+(2n-1)=
n(1+2n-1)
2
=n2
(n≥2)

1
a1
=1也适合上式,

an=

1
n2

(2)证明:∵an=

1
n2

Sn=a1+a2+…+an=1+

1
22
+
1
32
+…+
1
n2

∵当n≥2时,

1
n2
1
(n-1)n
=
1
n-1
-
1
n

1+

1
22
+
1
32
+…+
1
n2
<1+[(1-
1
2
)+(
1
2
-
1
3
)+…+(
1
n-1
-
1
n+1
)]=2-
1
n+1
<2.

又∵

1
n2
1
n(n+1)
=
1
n
-
1
n+1

Sn>(1-

1
2
)+(
1
2
-
1
3
)+…+(
1
n
-
1
n+1
)=1-
1
n+1
=
n
n+1

∴当n≥2时,

n
n+1
Sn<2.

(3)∵

1
n2
=
4
4n2
4
(2n-1)(2n+1)
=2(
1
2n-1
-
1
2n+1
)

1+

1
22
+
1
32
+…+
1
n2
<1+2[(
1
3
-
1
5
)+(
1
5
-
1
7
)+…+(
1
2n-1
-
1
2n+1
)]

=

5
3
-
2
2n+1
5
3

当n≥2时,要Sn

6n
(n+1)(2n+1)
只需
n
n+1
6n
(n+1)(2n+1)

即需2n+1>6,显然这在n≥3时成立

S2=1+

1
4
=
5
4
,当n≥2时
6n
(n+1)(2n+1)
=
6×2
(2+1)(4+1)
=
4
5
显然
5
4
4
5

即当n≥2时Sn

6n
(n+1)(2n+1)
也成立

综上所述:当n≥2时,有

6n
(n+1)(2n+1)
Sn
5
3

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