问题 填空题
设函数f(x)=
2x
2x+
2
的图象上两点P1(x1,y1)、P2(x2,y2),若
OP
=
1
2
OP1
+
OP2
),且点P的横坐标为
1
2

(1)求证:P点的纵坐标为定值,并求出这个定值;
(2)求Sn=f(
1
n
)+f(
2
n
)+A+f(
n-1
n
)+f(
n
n

(3)记Tn为数列{
1
(Sn+
2
)(Sn+1+
2
)
}的前n项和,若Tn<a(Sn+1+
2
)对一切n∈N*都成立,试求a的取值范围.
答案

(1)证:∵

OP
=
1
2
OP1
+
OP2
),

∴P是P1P2的中点⇒x1+x2=1------(2分)

∴y1+y2=f(x1)+f(x2)=

2x1
2x1+
2
+
2x2
2x2+
2
=
2x1
2x1+
2
+
21-x1
21-x1+
2
=
2x1
2x1+
2
+
2
2
2x1+2
=1.

yp=

1
2
(y1+y2)=
1
2
..-----------------------------(4分)

(2)由(1)知x1+x2=1,f (x1)+f (x2)=y1+y2=1,f (1)=2-

2

Sn=f(

1
n
)+f(
2
n
)+…+f(
n-1
n
)+f(
n
n
),

Sn=f(

n
n
)+f(
n-1
n
)+…+f(
2
n
)+f(
1
n
),

相加得 2Sn=f(1)+[f(

1
n
)+f(
n-1
n
)]+[f(
2
n
)+f(
n-2
n
)]+…+[f(
n-1
n
)+f(
1
n
)]+f(1),

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

2

Sn=

n+3-2
2
2
.------------(8分)

(3)

1
(Sn+
2
)(Sn+1+
2
)
=
1
n+3
2
n+4
2
=
4
(n+3)(n+4)
=4(
1
n+3
-
1
n+4
)

Tn=4[(

1
4
-
1
5
)+(
1
5
-
1
6
)+…+(
1
n+3
-
1
n+4
)]--------------------(10分) 

Tn<a(Sn+1+

2
)⇔a
Tn
Sn+1+
2
=
2n
(n+4)2
=
2
n+
16
n
+8

n+

16
n
≥8,当且仅当n=4时,取“=”

2
n+
16
n
+8
2
8+8
=
1
8
,因此,a
1
8
-------------------(12分)

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