Question

已知等差数列{an}满足a2=0,a6+a8=-10,求数列an/(2^(n-1))的值

Answer 1

设数列{an/(2^(n-1)}的前n项和为Sn

所以 Sn=a1+a2/2+a3/4+...+an/2^(n-1)①

n=1时,Sn=S1=1

①式*1/2得,Sn/2=a1/2+a2/4+a3/8+...+an/2^n

所以n>1时,Sn/2=a1+（a2-a1）/2+...+（an-an-1)/2^(n-1)-an/2^n (注：an-1中n-1为下标)

=1-（1/2+1/4+...+1/2^(n-1)-(2-n)/2^n）=1-(1-1/2^（n-1）)-(2-n)/2^n

所以Sn=n/2^(n-1)