设数列{an}的前n项和为sn,已知sn=2an-2^(n+1),（1）.求证数列{an/2^n}为等差数列,并求{an}的通项公式；（2）设bn=logan/(n+1)2,数列{bn}的前n项和为Bn,若存在整数m,对任意n属于正整数且n>=2,都有B3n-Bn>m/20

Question

设数列{an}的前n项和为sn,已知sn=2an-2^(n+1),（1）.求证数列{an/2^n}为等差数列,并求{an}的通项公式；
（2）设bn=logan/(n+1) 2,数列{bn}的前n项和为Bn,若存在整数m,对任意n属于正整数且n>=2,都有B3n-Bn>m/20成立,求m的最大值

Answer 1

(1)
an=sn-s(n-1)=2an-2^(n+1)-2a(n-1)+2^n
an=2a(n-1)+2^n
an/2^n=a(n-1)/2^(n-1)+1
a1=2a1-2^2 a1=4 a1/2=4/2=2
所以{an/2^n}是已2为首项,1为公差的等差数列
an/2^n=2+(n-1)=n+1
an=(n+1)*2^n
(2)
bn=log2^n 2=1/n
B3n-Bn=1/(n+1)+1/(n+2)+...+1/3n>2n/3n=2/3
m/20