设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).(1)求数列{an}的通项公式;(2)设bn=logann+12,数列{bn}的前n项和为Bn,若存在整数m,使对任意n∈N*且

题目简介

设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).(1)求数列{an}的通项公式;(2)设bn=logann+12,数列{bn}的前n项和为Bn,若存在整数m,使对任意n∈N*且

题目详情

设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).
(1)求数列{an}的通项公式;
(2)设bn=log
an
n+1
2
,数列{bn}的前n项和为Bn,若存在整数m,使对任意n∈N*且n≥2,都有B3n-Bn
m
20
成立,求m的最大值;
题型:解答题难度:中档来源:不详

答案

(1)由Sn=2an-2n+1,得Sn-1=2an-1-2n(n≥2).
两式相减,得an=2an-2an-1-2n,即an-2an-1=2n(n≥2).
于是
an
2n
-
an-1
2n-1
=1
,所以数列{
an
2n
}
是公差为1的等差数列.
又S1=2a1-22,所以a1=4.
所以
an
2n
=2+(n-1)=n+1

故an=(n+1)•2n.

(2)因为bn=log
an
n+1
2=log2n2
=class="stub"1
n
,则B3n-Bn=class="stub"1
n+1
+class="stub"1
n+2
++class="stub"1
3n

f(n)=class="stub"1
n+1
+class="stub"1
n+2
++class="stub"1
3n
,则f(n+1)=class="stub"1
n+2
+class="stub"1
n+3
++class="stub"1
3n
+class="stub"1
3n+1
+class="stub"1
3n+2
+class="stub"1
3n+3

所以f(n+1)-f(n)=class="stub"1
3n+1
+class="stub"1
3n+2
+class="stub"1
3n+3
-class="stub"1
n+1
=class="stub"1
3n+1
+class="stub"1
3n+2
-class="stub"2
3n+3
>class="stub"1
3n+3
+class="stub"1
3n+3
-class="stub"2
3n+3
=0

即f(n+1)>f(n),所以数列{f(n)}为递增数列.
所以当n≥2时,f(n)的最小值为f(2)=class="stub"1
3
+class="stub"1
4
+class="stub"1
5
+class="stub"1
6
=class="stub"19
20

据题意,class="stub"m
20
<class="stub"19
20
,即m<19.又m为整数,
故m的最大值为18.

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