设正项数列{an}的前n项和为Sn,满足Sn=n2.(1)求{an}的通项公式;(2)设bn=1(an+1)(an+1+1),求数列{bn}的前n项的和Tn.(3)是否存在自然数m,使得m-24<Tn

题目简介

设正项数列{an}的前n项和为Sn,满足Sn=n2.(1)求{an}的通项公式;(2)设bn=1(an+1)(an+1+1),求数列{bn}的前n项的和Tn.(3)是否存在自然数m,使得m-24<Tn

题目详情

设正项数列{an}的前n项和为Sn,满足Sn=n2
(1)求{an}的通项公式;
(2)设bn=
1
(an+1)(an+1+1)
,求数列{bn}的前n项的和Tn
(3)是否存在自然数m,使得
m-2
4
<Tn
m
5
对一切n∈N*恒成立?若存在,求出m的值;若不存在,说明理由.
题型:解答题难度:中档来源:不详

答案

(1)∵Sn=n2,∴当n=1时,a1=S1=1;
当n≥2时,an=Sn-Sn-1=2n-1
a1=1满足上式,∴an=2n-1;
(2)由bn=class="stub"1
(an+1)(an+1+1)
=class="stub"1
4
•class="stub"1
n(n+1)
=class="stub"1
4
(class="stub"1
n
-class="stub"1
n+1
)

∴Tn=class="stub"1
4
(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
)=class="stub"1
4
(1-class="stub"1
n+1
)=class="stub"n
4(n+1)

(3)Tn+1-Tn=class="stub"n+1
4(n+2)
-class="stub"n
4(n+1)
=class="stub"1
4(n+1)(n+2)
>0,∴{Tn}单调递增,∴Tn≥T1=class="stub"1
8

∵Tn=class="stub"1
4
(1-class="stub"1
n+1
)<class="stub"1
4
,∴class="stub"1
8
≤Tn<class="stub"1
4

使得class="stub"m-2
4
<Tn<class="stub"m
5
对一切n∈N*恒成立,则
class="stub"1
4
≤class="stub"m
5
class="stub"m-2
4
<class="stub"1
8

class="stub"5
4
≤m<class="stub"5
2

∵m是自然数,∴m=2.

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