已知数列{an}是等差数列,a2=3,a4+a5+a6=27,Sn为数列{an}的前n项和(1)求an和Sn;(2)若bn=2an+1an,求数列{bn}的前n项和Tn.-数学

题目简介

已知数列{an}是等差数列,a2=3,a4+a5+a6=27,Sn为数列{an}的前n项和(1)求an和Sn;(2)若bn=2an+1an,求数列{bn}的前n项和Tn.-数学

题目详情

已知数列{an}是等差数列,a2=3,a4+a5+a6=27,Sn为数列{an}的前n项和
(1)求an和Sn;      
(2)若bn=
2
an+1an
,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

(1)由已知a4+a5+a6=27,可得3a5=27,
解得a5=9.(1分)
设等差数列{an}的公差为d,则a5-a2=3d=6,解得d=2..(2分)
∴an=a2+(n-2)d=3+(n-2)×2=2n-1,(4分)
sn=
n(a1+an)
2
=
n(1+2n-1)
2
=n2

综上,an=2n-1,sn=n2(7分)
(2)把an=2n-1代入得bn=class="stub"2
an+1an
=class="stub"2
(2n+1)(2n-1)
=class="stub"1
2n-1
-class="stub"1
2n+1

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

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