已知函数f(x)=x2x+1,数列{an}满足a1=f(1),an+1=f(an)(n∈N*).(Ⅰ)求a1,a2的值;(Ⅱ)求数列{an}的通项公式;(Ⅲ)设bn=an•an+1,求数列{bn}的前

题目简介

已知函数f(x)=x2x+1,数列{an}满足a1=f(1),an+1=f(an)(n∈N*).(Ⅰ)求a1,a2的值;(Ⅱ)求数列{an}的通项公式;(Ⅲ)设bn=an•an+1,求数列{bn}的前

题目详情

已知函数f(x)=
x
2x+1
,数列{an}满足a1=f(1),an+1=f(an)(n∈N*).
(Ⅰ)求a1,a2的值;
(Ⅱ)求数列{an}的通项公式;
(Ⅲ)设bn=an•an+1,求数列{bn}的前n项和Sn,并比较Sn
n
2n+18
题型:解答题难度:中档来源:北京模拟

答案

(Ⅰ)a1=f(1)=class="stub"1
2+1
=class="stub"1
3
,a2=f(a1)=f(class="stub"1
3
)=
class="stub"1
3
class="stub"2
3
+1
=class="stub"1
5

(Ⅱ)∵an+1=
an
2an+1

class="stub"1
an+1
=
2an+1
an
=2+class="stub"1
an

class="stub"1
an+1
-class="stub"1
an
=2

∵a1=class="stub"1
3
,∴class="stub"1
a1
=3
∴数列{class="stub"1
an
}
是首项为3,公差为2的等差数列,
class="stub"1
an
=2n+1

an=class="stub"1
2n+1

(Ⅲ)bn=anan+1=class="stub"1
(2n+1)(2n+3)
=class="stub"1
2
(class="stub"1
2n+1
-class="stub"1
2n+3
)

Sn=class="stub"1
2
(class="stub"1
3
-class="stub"1
5
+class="stub"1
5
-class="stub"1
7
+…+class="stub"1
2n+1
-class="stub"1
2n+3
)=class="stub"n
6n+9

n=1时,S1=class="stub"1
15
class="stub"n
2n+18
=class="stub"1
20
,Sn大于class="stub"n
2n+18

n=2时,S2=class="stub"2
21
class="stub"n
2n+18
=class="stub"1
11
,Sn大于class="stub"n
2n+18

n=3时,S3=class="stub"1
9
class="stub"n
2n+18
=class="stub"3
26
,Sn小于class="stub"n
2n+18

n=4时,S4=class="stub"4
33
class="stub"n
2n+18
=class="stub"2
17
,Sn大于class="stub"n
2n+18

猜想n≥4时,Sn大于class="stub"n
2n+18

证明如下:①n=4时,S4=class="stub"4
33
class="stub"n
2n+18
=class="stub"2
17
,Sn大于class="stub"n
2n+18
,结论成立;
②假设n=k时,结论成立,即class="stub"k
6k+9
>class="stub"k
2k+18
,∴2k>6k-9
n=k+1时,有2k+1+18>2(6k-9)+18>6(k+1)+9,
class="stub"k+1
6(k+1)+9
>class="stub"k+1
2k+1+18
,结论成立
由①②可知,结论成立.

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