已知正项数列{an}的前n项和为Sn,且an和Sn满足:4Sn=(an+1)2(n=1,2,3…),(1)求{an}的通项公式;(2)设bn=1an•an+1,求{bn}的前n项和Tn;(3)在(2)

题目简介

已知正项数列{an}的前n项和为Sn,且an和Sn满足:4Sn=(an+1)2(n=1,2,3…),(1)求{an}的通项公式;(2)设bn=1an•an+1,求{bn}的前n项和Tn;(3)在(2)

题目详情

已知正项数列{an}的前n项和为Sn,且an和Sn满足:4Sn=(an+1)2(n=1,2,3…),
(1)求{an}的通项公式;
(2)设bn=
1
anan+1
,求{bn}的前n项和Tn
(3)在(2)的条件下,对任意n∈N*,Tn
m
23
都成立,求整数m的最大值.
题型:解答题难度:中档来源:不详

答案

(1)∵4Sn=(an+1)2,①
∴4Sn-1=(an-1+1)2(n≥2),②
①-②得
4(Sn-Sn-1)=(an+1)2-(an-1+1)2.
∴4an=(an+1)2-(an-1+1)2.
化简得(an+an-1)•(an-an-1-2)=0.
∵an>0,∴an-an-1=2(n≥2).
∴{an}是以1为首项,2为公差的等差数列.
∴an=1+(n-1)•2=2n-1.
(2)bn=class="stub"1
an•an+1
=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
class="stub"1
2n-1
-class="stub"1
2n+1
).
∴Tn=class="stub"1
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"1
2
(1-class="stub"1
2n+1
)=class="stub"n
2n+1

(3)由(2)知Tn=class="stub"1
2
(1-class="stub"1
2n+1
),
Tn+1-Tn=class="stub"1
2
(1-class="stub"1
2n+3
)-class="stub"1
2
(1-class="stub"1
2n+1

=class="stub"1
2
class="stub"1
2n+1
-class="stub"1
2n+3
)>0.
∴数列{Tn}是递增数列.
∴[Tn]min=T1=class="stub"1
3

class="stub"m
23
class="stub"1
3

∴m<class="stub"23
3

∴整数m的最大值是7.

更多内容推荐