问题
解答题
设函数y=f(x)=
(1)求P点的纵坐标; (2)若Sn=f(
(3)记Tn为数列{
|
答案
(1)∵
=OP
(1 2
+OP1
),∴P为P1P2的中点,∴x1+x2=1OP2
∴y1+y2=
+2x1 2x1+ 2
=12x2 2x2+ 2
∴P的纵坐标为
;1 2
(2)由(1)知,x1+x2=1,y1+y2=1,f(1)=2-2
∵Sn=f(
)+f(1 n
)+…+f(2 n
)+f(n-1 n
),Sn=f(n n
)+f(n n
)+…+f(n-1 n
)+f(2 n
)1 n
∴2Sn=(n-1)+2(2-
)=n+3-22 2
∴Sn=
;n+3-2 2 2
(3)Sn+
=2
,Sn+1+n+3 2
=2 n+4 2
∴
=1 (Sn+
)(Sn+1+2
)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
)=1 n+4 n n+4
∵Tn<a(Sn+2+
)对一切n∈N*都成立2
∴a>
=Tn Sn+2+ 2 2 n+
+920 n
设g(n)=n+
,则g(n)在[20 n
,+∞)上是增函数,在(0,20
)上是减函数20
∴g(n)的最小值为9
∴
<2 n+
+920 n 1 9
∴a>
.1 9