数列an中,a1=1,an+1=2an+2的n次方,设bn=an／2∧n-1,证明bn是等差数列,求数列an的前n项和sn

题目

数列an中,a1=1,an+1=2an+2的n次方,设bn=an／2∧n-1,证明bn是等差数列,求数列an的前n项和sn

答案

a(n+1)=2an+2^na(n+1)/2^n=2an/2^n+1a(n+1)/2^n=an/2^(n-1)+1a(n+1)/2^n-an/2^(n-1)=1,为定值.a1/2^(1-1)=1/1=1数列{an/2^(n-1)}是以1为首项,1为公差的等差数列.bn=an/2^(n-1)数列{bn}是以1为首项,1为公差的等差数列...