题目内容
(2006•南京二模)如图,已知曲线C:y=
,Cn:y=
(n∈N*).从C上的点Qn(xn,yn)作x轴的垂线,交Cn于点Pn,再从点Pn作y轴的垂线,交C于点Qn+1(xn+1,yn+1),设x1=1,an=xn+1-xn,bn=yn-yn+1.
(Ⅰ)求Q1,Q2的坐标;
(Ⅱ)求数列{an}的通项公式;
(Ⅲ)记数列{an•bn}的前n项和为Sn,求证:Sn<
.
1 |
x |
1 |
x+2-n |
(Ⅰ)求Q1,Q2的坐标;
(Ⅱ)求数列{an}的通项公式;
(Ⅲ)记数列{an•bn}的前n项和为Sn,求证:Sn<
1 |
3 |
分析:(I)由Qn(xn,yn),Qn+1(xn+1,yn+1),知点Pn的坐标为(xn,yn+1),由此能求出点Q1、Q2的坐标;
(II)由Qn,Qn+1在曲线C上,知yn=
,yn+1=
,由Pn在曲线Cn上,知yn+1=
,由此能求出数列{an} 的通项公式;
(III)由xn=(xn-xn-1)+(xn-1-xn-2)+…+(x2-x1)+x1=2-(n-1)+2-(n-2)+…+2-1+1=1-
=2-21-n,知an•bn=(xn+1-xn)•(yn-yn+1)=2-n(
-
)=
(
-
)=
,由此入手能够证明sn<
.
(II)由Qn,Qn+1在曲线C上,知yn=
1 |
xn |
1 |
xn+1 |
1 |
xn+2-n |
(III)由xn=(xn-xn-1)+(xn-1-xn-2)+…+(x2-x1)+x1=2-(n-1)+2-(n-2)+…+2-1+1=1-
1-(
| ||
1-
|
1 |
xn |
1 |
xn+1 |
1 |
2n |
1 |
2-21-n |
1 |
2-2-n |
1 |
(2•2n-2)• (2•2n-1) |
1 |
3 |
解答:(I)解:∵Qn(xn,yn),Qn+1(xn+1,yn+1),
∴点Pn的坐标为(xn,yn+1)
∵x1=1∴y1=1,∴Q1(x1,y1)即Q1(1,1)
C1:y=
,令x=1则y2=
∴P1的坐标为(x1,y2)即(1,
)
令
=
得x2=
∴Q2(x2,y2)即Q1(
,
).-----------------------------------(2分)
(II)解:∵Qn,Qn+1在曲线C上,
∴yn=
,yn+1=
,
又∵Pn在曲线Cn上,
∴yn+1=
,--------------------------------(4分)
∴xn+1=xn+2-n,
∴an=2-n.-----------------------------------------(6分)
(III)证明:xn=(xn-xn-1)+(xn-1-xn-2)+…+(x2-x1)+x1
=2-(n-1)+2-(n-2)+…+2-1+1
=1-
=2-21-n.-------------------(9分)
∴an•bn=(xn+1-xn)•(yn-yn+1)=2-n(
-
)=
(
-
)=
,
∵2•2n-2≥2n,2•2n-1≥3,
∴an•bn≤
.--------------------------------(12分)
∴Sn=a1b1+a2b2+…+anbn
≤
+
+…+
=
•
=
(1-
)<
-----------------------(14分)
∴点Pn的坐标为(xn,yn+1)
∵x1=1∴y1=1,∴Q1(x1,y1)即Q1(1,1)
C1:y=
1 |
x+2-1 |
2 |
3 |
∴P1的坐标为(x1,y2)即(1,
2 |
3 |
令
2 |
3 |
1 |
x2 |
3 |
2 |
∴Q2(x2,y2)即Q1(
3 |
2 |
2 |
3 |
(II)解:∵Qn,Qn+1在曲线C上,
∴yn=
1 |
xn |
1 |
xn+1 |
又∵Pn在曲线Cn上,
∴yn+1=
1 |
xn+2-n |
∴xn+1=xn+2-n,
∴an=2-n.-----------------------------------------(6分)
(III)证明:xn=(xn-xn-1)+(xn-1-xn-2)+…+(x2-x1)+x1
=2-(n-1)+2-(n-2)+…+2-1+1
=1-
1-(
| ||
1-
|
=2-21-n.-------------------(9分)
∴an•bn=(xn+1-xn)•(yn-yn+1)=2-n(
1 |
xn |
1 |
xn+1 |
1 |
2n |
1 |
2-21-n |
1 |
2-2-n |
1 |
(2•2n-2)• (2•2n-1) |
∵2•2n-2≥2n,2•2n-1≥3,
∴an•bn≤
1 |
3•2n |
∴Sn=a1b1+a2b2+…+anbn
≤
1 |
3×2 |
1 |
3×22 |
1 |
3×2n |
1 |
6 |
1-(
| ||
1-
|
1 |
3 |
1 |
2n |
1 |
3 |
点评:本题主要考查了等比数列的通项公式,以及等比数列的求和公式和数列与不等式的综合,同时考查了转化的思想和计算的能力,属于难题.
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