等比数列 {an}的通项公式为an=2^(n-1),数列{bn}的通项公式为bn=(n+1)/(4an)(n∈N+）求数列{bn}的前n项和
发布网友
发布时间：2022-05-29 08:22
共2个回答
热心网友
时间：2023-11-03 16:03
Sn= b1+b2+...+bs=2/2^2+3/2^3+4/2^4...+（n+1）/2^n+1 1
1/2* Sn= 2/2^3+3/2^4+...+n/2^n+1+（n+1）/2^n+2 2
1 - 2 : 1/2* Sn=1/2+1/2^3+...+1/2^n+1-（n+1）/2^(n+2)
=1/2+1/4-1/2^(n+1)-（n+1）/2^(n+2)
热心网友
时间：2023-11-03 16:04
Tn=b1+b2+b3+……+bn
=(1+1)/2^(1+1)+(2+1)/2^(2+1)+(3+1)/2^(3+1)+……+(n+1)/2^(n+1)
=2/2^2+3/2^3+4/2^4+……+(n+1)/2^(n+1) ①
2Tn=2*[2/2^2+3/2^3+4/2^4+……+(n+1)/2^(n+1)]
=2/2^1+3/2^2+4/2^3+……+(n+1)/2^n ②
②-①
=2Tn-Tn
=【2/2^1+3/2^2+4/2^3+……+(n+1)/2^n 】-【2/2^2+3/2^3+4/2^4+……+(n+1)/2^(n+1)】
=2/2^1+1/2^2+1/2^3+……+1/2^n-(n+1)/2^(n+1)
=1+（1/2^2+1/2^3+……+1/2^n）-(n+1)/2^(n+1)
=1+1/4*(1-1/2^（n-1)/(1-1/2) -(n+1)/2^(n+1)
=1+1/8-1/2^（n+2)-(2n+2)/2^(n+2)
=9/8-(2n+3)/2^(n+2)
