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,nn+1<Sn<2;(3

题目简介

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,nn+1<Sn<2;(3

题目详情

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.
题型:解答题难度:中档来源:不详

答案

(1)∵an≠0
∴anan-1≠0(n≥2)
an
anan-1
=
(1-2n)anan-1
anan-1
+
an-1
anan-1

class="stub"1
an-1
=(1-2n)+class="stub"1
an
即有class="stub"1
an
-class="stub"1
an-1
=2n-1

class="stub"1
an
=class="stub"1
a1
+(class="stub"1
a2
-class="stub"1
a1
)+(class="stub"1
a3
-class="stub"1
a2
)+…+(class="stub"1
an
-class="stub"1
an-1
)
=1+3+5+7+…+(2n-1)=
n(1+2n-1)
2
=n2
(n≥2)
class="stub"1
a1
=1
也适合上式,
an=class="stub"1
n2

(2)证明:∵an=class="stub"1
n2

Sn=a1+a2+…+an=1+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2

∵当n≥2时,class="stub"1
n2
<class="stub"1
(n-1)n
=class="stub"1
n-1
-class="stub"1
n

1+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
<1+[(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n-1
-class="stub"1
n+1
)]
=2-class="stub"1
n+1
<2.
又∵class="stub"1
n2
>class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

Sn>(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1
)
=1-class="stub"1
n+1
=class="stub"n
n+1

∴当n≥2时,class="stub"n
n+1
Sn<2

(3)∵class="stub"1
n2
=class="stub"4
4n2
<class="stub"4
(2n-1)(2n+1)
=2(class="stub"1
2n-1
-class="stub"1
2n+1
)

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

=class="stub"5
3
-class="stub"2
2n+1
<class="stub"5
3

当n≥2时,要Sn>class="stub"6n
(n+1)(2n+1)
只需class="stub"n
n+1
>class="stub"6n
(n+1)(2n+1)

即需2n+1>6,显然这在n≥3时成立
S2=1+class="stub"1
4
=class="stub"5
4
,当n≥2时class="stub"6n
(n+1)(2n+1)
=class="stub"6×2
(2+1)(4+1)
=class="stub"4
5
显然class="stub"5
4
>class="stub"4
5

即当n≥2时Sn>class="stub"6n
(n+1)(2n+1)
也成立
综上所述:当n≥2时,有class="stub"6n
(n+1)(2n+1)
Sn<class="stub"5
3

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