问题 解答题
已知函数y=f(x)满足
a
=(x2,y),
b
=(x-
1
x
,-1)
,且
a
b
=-1

如果存在正项数列{an}满足:a1=
1
2
 f(a1)+f(a2)+f(a3)+…+f(an)-n
=a13+a23+a33+…+an3-n2an(n∈N*).
(1)求数列{an}的通项;
(2)求证:
a1
1
+
a2
2
+
a3
3
+…+
an
n
<1

(3)求证:
a1
1
+
a2
2
+
a3
3
+…+
an
n
<1+
2
答案

(1)

a
b
=-1,∴y=f(x)=x3-x+1(x≠0)

∵f(a1)+f(a2)+f(a3)+…+f(an)-n=a13+a23+a33+…+an3-n2an(n∈N*).

所以代入得a1+a2+a3+…+an=n2an

又a1+a2+a3+…+an-1=(n-1)2an-1(n≥2)②

①-②得 

an
an-1
=
n-1
n+1
an=
an
an-1
an-1
an-2
•…•
a2
a1
=
1
n(n+1)
(n∈N*)
…(4分)

(2)由(1)得

ai
i
=
1
i2(i+1)
=
1
i
(
1
i
-
1
i+1
)=
1
i2
-(
1
i
-
1
i+1
)
1
i(i-1)
-(
1
i
-
1
i+1
)
=(
1
i-1
-
1
i
)+(
1
i
-
1
i+1
)(i>1)

a1
1
+
a2
2
+
a3
3
+…+
an
n
1
2
+
1
2
-(
1
n
-
1
n+1
)=1+
1
n+1
-
1
n
=1-
1
n(n+1)
<1…(9分)

(3)∵

i+1
+
i-1
2
(i+1)+(i-1)
2
=
i
1
i
2
i+1
+
i-1

ai
i
=
1
i2(i+1)
1
(i-1)i(i+1)
1
i-1
i+1
2
i-1
+
i+1
=
1
i-1
-
1
i+1
(i≥2)

所以

a1
1
+
a2
2
+
a3
3
+…+
an
n
a1
1
+(
1
1
-
1
3
)+(
1
2
-
1
4
)+…+(
1
n-1
-
1
n+1
)<
1
2
+1+
1
2
-
1
n
-
1
n+1
<1+
2
…(14分)

单项选择题
单项选择题