解答:(I)解:∵Q
n(x
n,y
n),Q
n+1(x
n+1,y
n+1),
∴点P
n的坐标为(x
n,y
n+1)
∵x
1=1∴y
1=1,∴Q
1(x
1,y
1)即Q
1(1,1)
C1:y=,令x=1则y
2=
∴P
1的坐标为(x
1,y
2)即(1,
)
令
=
得x
2=
∴Q
2(x
2,y
2)即Q
1(
,
).-----------------------------------(2分)
(II)解:∵Q
n,Q
n+1在曲线C上,
∴
yn=,
yn+1=,
又∵P
n在曲线C
n上,
∴
yn+1=,--------------------------------(4分)
∴x
n+1=x
n+2
-n,
∴a
n=2
-n.-----------------------------------------(6分)
(III)证明:x
n=(x
n-x
n-1)+(x
n-1-x
n-2)+…+(x
2-x
1)+x
1=2
-(n-1)+2
-(n-2)+…+2
-1+1
=
1-=2-2
1-n.-------------------(9分)
∴a
n•b
n=(x
n+1-x
n)•(y
n-y
n+1)=
2-n(-)=
(-)=
,
∵2•2
n-2≥2
n,2•2
n-1≥3,
∴
an•bn≤.--------------------------------(12分)
∴S
n=a
1b
1+a
2b
2+…+a
nb
n≤++…+=•=(1-)<-----------------------(14分)