已知函数f(x)=2x+33x,数列{an}满足a1=1,an+1=f(1an),n∈N*.(1)求数列{an}的通项公式;(2)令Tn=a1a2-a2a3+a3a4-a4a5+…+a2n-1a2n-

题目简介

已知函数f(x)=2x+33x,数列{an}满足a1=1,an+1=f(1an),n∈N*.(1)求数列{an}的通项公式;(2)令Tn=a1a2-a2a3+a3a4-a4a5+…+a2n-1a2n-

题目详情

已知函数f(x)=
2x+3
3x
,数列{an}满足a1=1,an+1=f(
1
an
),n∈N*

(1)求数列{an}的通项公式;
(2)令Tn=a1a2-a2a3+a3a4-a4a5+…+a2n-1a2n-a2na2n+1,求Tn
(3)令bn=
1
an-1an
(n≥2)
,b1=3,Sn=b1+b2+…+bn,若Sn
m-2002
2
对一切n∈N*成立,求最小正整数m.
题型:解答题难度:中档来源:韶关模拟

答案

(1)∵an+1=f(class="stub"1
an
)=
2+3an
3
=an+class="stub"2
3

an+1-an=class="stub"2
3

∴数列{an}是以class="stub"2
3
为公差,首项a1=1的等差数列
an=class="stub"2
3
n+class="stub"1
3

(2)Tn=a1a2-a2a3+a3a4-a4a5+…+a2n-1a2n-a2na2n+1
=a2(a1-a3)+a4(a3-a5)+…+a2n(a2n-1-a2n+1)
=-class="stub"4
3
(a2+a4+…+a2n)

=-class="stub"4
3
×
n(class="stub"5
3
+class="stub"4n
3
+class="stub"1
3
)  
2

=-class="stub"4
9
(2n2+3n)

(3)当n≥2时,bn=class="stub"1
an-1an
=class="stub"1
(class="stub"2
3
n-class="stub"1
3
)(class="stub"2
3
n+class="stub"1
3
)
=class="stub"9
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)

当n=1时,上式同样成立
∴sn=b1+b2+…+bn=class="stub"9
2
[(1-class="stub"1
3
) +(class="stub"1
3
-class="stub"1
5
)+…+(class="stub"1
2n-1
-class="stub"1
2n+1
)]

=class="stub"9
2
(1-class="stub"1
2n+1
)

∵恒有class="stub"9
2
(1-class="stub"1
2n+1
)<class="stub"9
2
成立,
Sn<class="stub"m-2002
2
,即class="stub"9
2
(1-class="stub"1
2n+1
)<class="stub"m-2002
2
对一切n∈N*成立,
class="stub"9
2
≤class="stub"m-2002
2
,解得  m≥2011,
∴m最小=2011

更多内容推荐