Question

已知数列An是等差数列且An=2^n-1,Bn是等比数列且Bn=2n-1，求数列Bn/An的前n项和Sn

Answer 1

应该是打错了吧 An等比 Bn等差吧
Bn/An=(2n-1)/2^n-1=(2n-1)2^(1-n)
Sn=(2*1-1)2^0+(2*2-1)2^(-1)+(2*3-1)2^(-2)……(2n-1)2^(1-n)
1/2Sn= (2*1-1)2^(-1)+(2*2-1)2^(-2)……(2n-3)2^(1-n)+(2n-1)2^(1-n)
两式相减得到
1/2Sn=1+2*2^(-1)+2*2^(-2)+……+2*2^(1-n)-(2n-1)2^(-n)
=1+1+1/2……+1/2^(n-2)-(2n-1)2^(-n)
=1+{1-(1/2)^(n-1)}/(1-1/2)-(2n-1)2^(-n)
=3-1/2^(n-2)-(2n-1)/2^n
Sn=6-1/2^(n-3)-(2n-1)/2^(n-1)