已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=2n-1an+1an,n∈N*(1)求证Sn=2n-1an(2)设bn=anan+1求数列{bn}的前n项和Tn.-数学

题目简介

已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=2n-1an+1an,n∈N*(1)求证Sn=2n-1an(2)设bn=anan+1求数列{bn}的前n项和Tn.-数学

题目详情

已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=2n-1an+1an,n∈N*
(1)求证Sn=2n-1an
(2)设bn=
an
an+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

(1)证明:∵an≠0,anSn+1-an+1Sn=2n-1an+1an,n∈N*
Sn+1
an+1
-
Sn
an
=2n-1
cn=
Sn
an
,则cn+1-cn=2n-1
利用叠加法可得:cn-c1=20+21+…+2n-2=
1-2n-1
1-2
=2n-1-1
c1=
S1
a1
=1,∴cn=2n-1
Sn=2n-1an
(2)由(1)知,Sn+1=2nan+1
两式相减可得an+1=2nan+1-2n-1an
bn=
an
an+1
=
2n-1
2n-1
=2(1-class="stub"1
2n
)

∴Tn=class="stub"n
2
-2(class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n
)
=class="stub"n-4
2
+class="stub"1
2n-1

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