正项数列{an}的前n项和为Sn,且2Sn=an+1.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=1an•an+1,数列{bn}的前n项和为Tn,求证:Tn<12.-数学

题目简介

正项数列{an}的前n项和为Sn,且2Sn=an+1.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=1an•an+1,数列{bn}的前n项和为Tn,求证:Tn<12.-数学

题目详情

正项数列{an}的前n项和为Sn,且2
Sn
=an+1

(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=
1
anan+1
,数列{bn}的前n项和为Tn,求证:Tn
1
2
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵2
S1
=a1+1

∴a1=1.
∵an>0,2
Sn
=an+1

∴4Sn=(an+1)2.①
∴4Sn-1=(an-1+1)2(n≥2).②
①-②,得4an=an2+2an-an-12-2an-1,
即(an+an-1)(an-an-1-2)=0,
而an>0,
∴an-an-1=2(n≥2).
故数列{an}是首项为1,公差为2的等差数列.
∴an=2n-1.
(Ⅱ)bn=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)

Tn=b1+b2++bn=class="stub"1
2
(1-class="stub"1
3
)+class="stub"1
2
(class="stub"1
3
-class="stub"1
5
)++class="stub"1
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)
=class="stub"1
2
(1-class="stub"1
2n+1
)<class="stub"1
2

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