设函数f(x)=ax2+bx+1x+c(a>0)为奇函数,且|f(x)|min=22,数列{an}与{bn}满足如下关系:a1=2,an+1=f(an)-an2,bn=an-1an+1.(1)求f(x

题目简介

设函数f(x)=ax2+bx+1x+c(a>0)为奇函数,且|f(x)|min=22,数列{an}与{bn}满足如下关系:a1=2,an+1=f(an)-an2,bn=an-1an+1.(1)求f(x

题目详情

设函数f(x)=
ax2+bx+1
x+c
(a>0)
为奇函数,且|f(x)|min=2
2
,数列{an}与{bn}满足如下关系:a1=2,an+1=
f(an)-an
2
bn=
an-1
an+1
.

(1)求f(x)的解析式;
(2)求数列{bn}的通项公式bn
(3)记Sn为数列{an}的前n项和,求证:对任意的n∈N*Sn<n+
3
2
.
题型:解答题难度:中档来源:不详

答案

(1)由f(x)是奇函数,得b=c=0,
|f(x)|min=2
2
,得a=2,故f(x)=
2x2+1
x
.

(2)∵an+1=
f(an)-an
2
=
a2n
+1
2an

bn+1=
an+1-1
an+1+1
=
a2n
+1
2an
-1
a2n
+1
2an
+1
=
a2n
-2an+1
a2n
+2an+1
=(
an-1
an+1
)2=
b2n

bn=
b2n-1
=
b4n-2
b2n-11

b1=class="stub"1
3
,∴bn=(class="stub"1
3
)2n-1

(3)证明:由(2)
an-1
an+1
=(class="stub"1
3
)2n-1an=
1+(class="stub"1
3
)
2n-1
1-(class="stub"1
3
)
2n-1
=
32n-1+1
32n-1-1
=1+class="stub"2
32n-1-1

要证明的问题即为class="stub"2
321-1-1
+class="stub"2
322-1-1
++class="stub"2
32n-1-1
<class="stub"3
2

当n=1时,2n-1=n
当n≥2时,2n-1=(1+1)n-1≥Cn-10+Cn-11=n∴2n-1≥n
32n-13n=3×3n-1=2×3n-1+3n-1≥2×3n-1+1
class="stub"2
32n-1-1
≤(class="stub"1
3
)n-1

class="stub"2
321-1-1
+class="stub"2
322-1-1
++class="stub"2
32n-1-1
≤1+class="stub"1
3
+(class="stub"1
3
)2++(class="stub"1
3
)n-1=
[1-(class="stub"1
3
)
n
]
1-class="stub"1
3

=class="stub"3
2
-class="stub"3
2
(class="stub"1
3
)n<class="stub"3
2
得证.

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