设函数f(x)=
(1)求证:P点的纵坐标为定值,并求出这个定值; (2)求Sn=f(
(3)记Tn为数列{
|
(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
=1.2
•2x1+22
∴yp=
(y1+y2)=1 2
..-----------------------------(4分)1 2
(2)由(1)知x1+x2=1,f (x1)+f (x2)=y1+y2=1,f (1)=2-
,2
Sn=f(
)+f(1 n
)+…+f(2 n
)+f(n-1 n
),n n
Sn=f(
)+f(n n
)+…+f(n-1 n
)+f(2 n
),1 n
相加得 2Sn=f(1)+[f(
)+f(1 n
)]+[f(n-1 n
)+f(2 n
)]+…+[f(n-2 n
)+f(n-1 n
)]+f(1),1 n
=2f(1)+n-1=n+3-22
∴Sn=
.------------(8分)n+3-2 2 2
(3)
=1 (Sn+
)(Sn+1+2
)2
=1
•n+3 2 n+4 2
=4(4 (n+3)(n+4)
-1 n+3
),1 n+4
Tn=4[(
-1 4
)+(1 5
-1 5
)+…+(1 6
-1 n+3
)]--------------------(10分) 1 n+4
Tn<a(Sn+1+
)⇔a>2
=Tn Sn+1+ 2
=2n (n+4)2 2 n+
+816 n
∵n+
≥8,当且仅当n=4时,取“=”16 n
∴
≤2 n+
+816 n
=2 8+8
,因此,a>1 8
-------------------(12分)1 8