设正数数列{an}的前n项和Sn满足Sn=14(an+1)2.(I)求数列{an}的通项公式;(II)设bn=1an•an+1,求数列{bn}的前n项和Tn.-数学

题目简介

设正数数列{an}的前n项和Sn满足Sn=14(an+1)2.(I)求数列{an}的通项公式;(II)设bn=1an•an+1,求数列{bn}的前n项和Tn.-数学

题目详情

设正数数列{an}的前n项和Sn满足Sn=
1
4
(an+1)2

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

答案

(Ⅰ)当n=1时,a1=S1=class="stub"1
4
(a1+1)2

∴a1=1.(2分)
Sn=class="stub"1
4
(an+1)2
,①
Sn-1=class="stub"1
4
(an-1+1)2
(n≥2).②
①-②,得an=Sn-Sn-1=class="stub"1
4
(an+1)2-class="stub"1
4
(an-1+1)2

整理得,(an+an-1)(an-an-1-2)=0,(5分)
∵an>0
∴an+an-1>0.
∴an-an-1-2=0,即an-an-1=2(n≥2).(7分)
故数列{an}是首项为1,公差为2的等差数列.
∴an=2n-1.(9分)
(Ⅱ)∵bn=class="stub"1
anan+1
=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)
,(11分)
∴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"n
2n+1
. (14分)

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