已知数列{an}中,a1=1,an<an+1,设bn=an+1-anan+1•an+1,Sn=b1+b2+…+bn,求证:(Ⅰ)bn<2(1an-1an+1);

已知数列{an}中,a1=1,an<an+1,设bn=an+1-anan+1•an+1,Sn=b1+b2+…+bn,求证:(Ⅰ)bn<2(1an-1an+1);

题型:不详难度:来源:
已知数列{an}中,a1=1,an<an+1,设bn=
an+1-an
an+1


an+1
,Sn=b1+b2+…+bn,求证:
(Ⅰ)bn<2(
1


an
-
1


an+1
)

(Ⅱ)若数列{an}是公比为q且q≥3的等比数列,则Sn<1.
答案
证明:(Ⅰ)由题意可知an>0
bn-2(
1


an
-
1


an+1
)

=
an+1-an
an+1


an+1
-2(
1


an
-
1


an+1
)

=
an+1-an
an+1


an+1
-2


an+1
-


an


an+1


an

=
(


an+1
-


an
)(


an+1


an
+an-2an+1)
an+1


an+1


an

又an<an+1,∴


an+1
-


an
>0


an+1


an
an+1



an+1


an
+an-2an+1<0

(


an+1
-


an
)(


an+1


an
+an-2an+1)
an+1


an+1


an
<0

bn<2(
1


an
-
1


an+1
)

(Ⅱ)数列{an}是首项a1=1,公比为q且q≥3的等比数列,
an=a1qn-1=qn-1
bn=
an+1-an
an+1


an+1
=
qn-qn-1
q
3n
2
=q-
n
2
(1-q-1)

Sn=b1+b2+…+bn
=(1-q-1)(q-
1
2
+q-
2
2
+q-
3
2
+…+q-
n
2
)

=(1-q-1)•
q-
1
2
(1-q-
n
2
)
1-q-
1
2

=(
1


q
+
1
q
)(1-
1


qn
)

∵q≥3,∴0<
1


q
+
1
q
1


3
+
1
3
=


3
+1
3
<1

0<1-
1


qn
<1

Sn=(
1


q
+
1
q
)(1-
1


qn
)<1
举一反三
给出集合序列{1},{2,3},{4,5,6},{7,8,9,10},…,设Sn是第n个集合中元素之和,则S21为(  )
A.1113B.4641C.5082D.5336
题型:不详难度:| 查看答案
数列{(-1)n•n}的前n项和为Sn,则S2007等于(  )
A.1004B.-1004C.2005D.-2005
题型:不详难度:| 查看答案
已知数列{an}的前n项和Sn满足:Sn+Sm=Sn+m,且a1=1,则a2013=______.
题型:不详难度:| 查看答案
已知数列{an}的通项为an,前n项和为sn,且an是sn与2的等差中项,数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=0上.
(Ⅰ)求数列{an}、{bn}的通项公式an,bn
(Ⅱ)设{bn}的前n项和为Bn,试比较
1
B1
+
1
B2
+…+
1
Bn
与2的大小.
(Ⅲ)设Tn=
b1
a1
+
b2
a2
+…+
bn
an
,若对一切正整数n,Tn<c(c∈Z)恒成立,求c的最小值.
题型:不详难度:| 查看答案
已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则
1
S1
+
1
S2
+
1
S3
+…+
1
Sn
=______.
题型:不详难度:| 查看答案
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