设数列{an}前n项和Sn,且Sn=2an-2,令bn=log2an(I)试求数列{an}的通项公式;(Ⅱ)设cn=bnan,求证数列{cn}的前n项和Tn<2.(Ⅲ)对任意m∈N*,将数列{2bn}

题目简介

设数列{an}前n项和Sn,且Sn=2an-2,令bn=log2an(I)试求数列{an}的通项公式;(Ⅱ)设cn=bnan,求证数列{cn}的前n项和Tn<2.(Ⅲ)对任意m∈N*,将数列{2bn}

题目详情

设数列{an}前n项和Sn,且Sn=2an-2,令bn=log2an
(I)试求数列{an}的通项公式;
(Ⅱ)设cn=
bn
an
,求证数列{cn}的前n项和Tn<2.
(Ⅲ)对任意m∈N*,将数列{2bn}中落入区间(am,a2m)内的项的个数记为dm,求数列{dm}的前m项和Tm
题型:解答题难度:中档来源:不详

答案

(I)当n=1时,S1=2a1-2,a1=2,
当n≥2时,an=Sn-Sn-1=(2an-2)-(2an-1-2)=2an-2an-1,所以an=2an-1,数列{an}是以2为为公比的等比数列,且首项a1=2,
通项公式为an=2×2n-1=2n,
(Ⅱ)cn=
bn
an
=class="stub"n
2n

Tn=class="stub"1
21
+class="stub"2
22
+…class="stub"n
2n
,两边同乘以class="stub"1
2

class="stub"1
2
Tn=class="stub"1
22
+class="stub"2
23
+…class="stub"n-1
2n
+class="stub"n
2n+1

两式相减得出class="stub"1
2
Tn=class="stub"1
21
+class="stub"1
22
+…class="stub"1
2n
-class="stub"n
2n+1
=1-class="stub"1
2n
-class="stub"n
2n+1
=1-class="stub"n+2
2n+1

∴Tn=2-class="stub"n+2
2n

∴Tn<2
(Ⅲ)数列{2bn}中落入区间(am,a2m)内,即am<2bn<a2m,所以2m<2n<22m,2m-1<n<22m-1,
所以数列{2bn}中落入区间(am,a2m)内的项的个数dm=22m-1-2m-1-1,
所以Tm.=
2(4m-1)
4-1
-
2m-1
2-1
-m
=class="stub"1
3
×22m+1-2m-m+class="stub"1
3

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