问题
解答题
已知函数y=f(x)满足
如果存在正项数列{an}满足:a1=
(1)求数列{an}的通项; (2)求证:
(3)求证:
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答案
(1)
•a
=-1,∴y=f(x)=x3-x+1(x≠0)b
∵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
则an=n-1 n+1
•an an-1
•…•an-1 an-2
=a2 a1
(n∈N*)…(4分)1 n(n+1)
(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
)(i>1)1 i+1
∴
+a1 1
+a2 2
+…+a3 3
<an n
+1 2
-(1 2
-1 n
)=1+1 n+1
-1 n+1
=1-1 n
<1…(9分)1 n(n+1)
(3)∵
<
+i+1 i-1 2
=(i+1)+(i-1) 2
∴i
<1 i 2
+i+1 i-1
而
=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
(i≥2)1 i+1
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+1 2
-1 2
-1 n
<1+1 n+1
…(14分)2