Question

对于等差数列除以等比数列前n项和的求解

设等差数列An=2-n，等比数列Bn=2^(n-1)，求数列An/Bn的前n项和。求助啊！

Answer 1

Sn=1*(1/2)^0+0*(1/2)^1 +(-1)*(1/2)^2+(-2)*(1/2)^3 +……+(2-n)(1/2)^(n-1)......................①
(1/2)Sn= 1*(1/2)^1 +0*(1/2)^2 +(-1)*(1/2)^3 +(-2)*(1/2)^4+……+(2-n)(1/2)^n........②
①-②
(1/2)Sn=1*(1/2)^0+(-1)*(1/2)^1 +(-1)*(1/2)^2+(-1)*(1/2)^3 +……+(-1)(1/2)^(n-1)+(n-2)(1/2)^n
=1-[ (1/2)^1 +(1/2)^2+(1/2)^3 +……+(1/2)^(n-1) ] + (n-2)(1/2)^n
=1-[1-2^(1-n)] + (n-2)/2^n
=1/2^(n-1) +(n/2 -1)/2^(n-1)
=(n/2)/2^(n-1)
=n/2^n
Sn=2n/2^n = n / 2^(n-1) 即：Sn = n / 2^(n-1)
祝你学习进步，更上一层楼！ (*^__^*)

Answer 2

可知：An/Bn=(2-n)/[2^(n-1)]=2^(2-n)-n*2^(1-n)
令Tn=An/Bn，an=2^(2-n)，bn=n*2^(1-n)
则Tn的前n项和即是an的前n项和减去bn的前n项和
令an、bn、Tn的前n项和分别为：S1、S2、Sn
1)、由an=2^(2-n)可知，an是首项为2，公比为1/2的等比数列
所以，根据等比数列的求和公式可求出S1
S1=2[1-(1/2)^n]/(1-1/2)=4-2^(2-n)
2）、由bn=n*2^(1-n)可知，
S2=1*2^0+2*2^(-1)+3*2^(-2)+4*2^(-3)+.......+(n-1)*2^(2-n)+n*2^(1-n)
(1/2)*S2=1*2^(-1)+2*2^(-2)+3*2^(-3)+......+(n-1)*2^(1-n)+n*2^(-n)
S2-(1/2)*S2=2^0+2^(-1)+2^(-2)+2^(-3)+.....+2^(1-n)-n*2^(-n)
=[1-(1/2)^n]/[1-1/2]-n*2^(-n)
=2-2^(1-n)-n*2^(-n)
=2-(2+n)*2^(-n)
即，S2=[2-(2+n)*2^(-n)]*2=4-(2+n)*2^(1-n)
则：Sn=S1-S2
=4-2^(2-n)-4-(2+n)*2^(1-n)
=(-4-n)*2^(1-n)
综上所述：An/Bn的前n项和为：(-4-n)*2^(1-n)

Answer 3

请参考下图：

Answer 4

Sn=1*(1/2)^0+0*(1/2)^1+(-1)*(1/2)^2+(-3)*(1/2)^3+……+(2-n)(1/2)^(n-1)
(1/2)Sn=1*(1/2)^1+0*(1/2)^2+(-1)*(1/2)^3+(-3)*(1/2)^4+……+(2-n)(1/2)^n
两式错位相减得到
(1/2)Sn=1*(1/2)^0+(-1)[(1/2)^1+(1/2)^2+(1/2)^3+……+(1/2)^(n-1)]-(2-n)(1/2)^n
=1-[(1/2-(1/2)^n)/(1-1/2)]-(2-n)(1/2)^n
=1-1+(1/2)^(n-1)-(2-n)^(1/2)n
=(1/2)^(n-1)-(2-n)^(1/2)n
所以Sn=(1/2)^(n-2)-(2-n)(1/2)^(n-1)=n*2^(1-n)

Answer 5

Sn=1*(1/2)^0+0*(1/2)^1 +(-1)*(1/2)^2+(-2)*(1/2)^3 +……+(2-n)(1/2)^(n-1)......................①
(1/2)Sn= 1*(1/2)^1 +0*(1/2)^2 +(-1)*(1/2)^3 +(-2)*(1/2)^4+……+(2-n)(1/2)^n........②
①-②
(1/2)Sn=1*(1/2)^0+(-1)*(1/2)^1 +(-1)*(1/2)^2+(-1)*(1/2)^3 +……+(-1)(1/2)^(n-1)+(n-2)(1/2)^n
=1-[ (1/2)^1 +(1/2)^2+(1/2)^3 +……+(1/2)^(n-1) ] + (n-2)(1/2)^n
=1-[1-2^(1-n)] + (n-2)/2^n
=1/2^(n-1) +(n/2 -1)/2^(n-1)
=(n/2)/2^(n-1)
=n/2^n
Sn=2n/2^n = n / 2^(n-1) 即：Sn = n / 2^(n-1)
祝你学习进步，更上一层楼！ (*^__^*)