已知函数y=f(x)满足a=(x2,y),b=(x-1x,-1),且a•b=-1.如果存在正项数列{an}满足:a1=12,f(a1)+f(a2)+f(a3)+…+f(an)-n=a13+a23+a3

题目简介

已知函数y=f(x)满足a=(x2,y),b=(x-1x,-1),且a•b=-1.如果存在正项数列{an}满足:a1=12,f(a1)+f(a2)+f(a3)+…+f(an)-n=a13+a23+a3

题目详情

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
题型:解答题难度:中档来源:不详

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 ①
又a1+a2+a3+…+an-1=(n-1)2an-1(n≥2)②
①-②得 
an
an-1
=class="stub"n-1
n+1
an=
an
an-1
an-1
an-2
•…•
a2
a1
=class="stub"1
n(n+1)
(n∈N*)
…(4分)
(2)由(1)得
ai
i
=class="stub"1
i2(i+1)
=class="stub"1
i
(class="stub"1
i
-class="stub"1
i+1
)=class="stub"1
i2
-(class="stub"1
i
-class="stub"1
i+1
)
<class="stub"1
i(i-1)
-(class="stub"1
i
-class="stub"1
i+1
)
=(class="stub"1
i-1
-class="stub"1
i
)+(class="stub"1
i
-class="stub"1
i+1
)(i>1)

a1
1
+
a2
2
+
a3
3
+…+
an
n
<class="stub"1
2
+class="stub"1
2
-(class="stub"1
n
-class="stub"1
n+1
)=1+class="stub"1
n+1
-class="stub"1
n
=1-class="stub"1
n(n+1)
<1
…(9分)
(3)∵
i+1
+
i-1
2
(i+1)+(i-1)
2
=
i
class="stub"1
i
<class="stub"2
i+1
+
i-1

ai
i
=
class="stub"1
i2(i+1)
class="stub"1
(i-1)i(i+1)
<class="stub"1
i-1
i+1
•class="stub"2
i-1
+
i+1
=class="stub"1
i-1
-class="stub"1
i+1
(i≥2)

所以
a1
1
+
a2
2
+
a3
3
+…+
an
n
a1
1
+(class="stub"1
1
-class="stub"1
3
)+(class="stub"1
2
-class="stub"1
4
)+…+
(class="stub"1
n-1
-class="stub"1
n+1
)<class="stub"1
2
+1+class="stub"1
2
-class="stub"1
n
-class="stub"1
n+1
<1+
2
…(14分)

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