一道有3问的数列题，谢谢啦
- 提问者网友：捧腹剧
- 2021-05-07 09:42
- 五星知识达人网友：西风乍起
- 2021-05-07 11:18
１、
2a(n+1)=(1+1/n)^2an
2a(n+1)=an(n+1)^2/n^2
2a(n+1)/(n+1)^2=an/n^2
[a(n+1)/(n+1)^2]/[an/n^2]=1/2
所以，数列 {an/n^2}后项与前项之比为１／２，是等比数列
an/n^2=a1/1^2*(1/2)^(n-1)
=(1/2)^(n-1)
an=n^2/2^(n-1)
２、
bn=a(n+1)-1/2an
=(n+1)^2/2^n-n^2/2^n
=1/2^n
Sn=1/2+1/4+..+1/2^n=1/2×[1-(1/2)^n/(1-1/2)]=1-1/2^n
3、
b1=a2-a1/2
b2=a3-a2/2
....
bn=an-a(n-1)/2
b1+b2+...+bn=-a1/2+a2/2+a3/2+..+a(n-1)/2+an
Sn=a1/2+a2/2+...+an/2+an/2-a1
Sn=Tn/2+n^2/2^n-1
2Sn=Tn+n^2/2^(n-1)-2
Tn=2Sn-n^2/2^(n-1)+2
=2(1-1/2^n)-n^2/2^(n-1)+2
=4-(n^2+1)/2^(n-1)
- 1楼网友：轮獄道
- 2021-05-07 11:43
１、
2a(n+1)=(1+1/n)^2an
2a(n+1)=an(n+1)^2/n^2
2a(n+1)/(n+1)^2=an/n^2
[a(n+1)/(n+1)^2]/[an/n^2]=1/2
所以，数列 {an/n^2}后项与前项之比为１／２，是等比数列
an/n^2=a1/1^2*(1/2)^(n-1)
=(1/2)^(n-1)
an=n^2/2^(n-1)
２、
bn=a(n+1)-1/2an
=(n+1)^2/2^n-n^2/2^n
=1/2^n
Sn=1/2+1/4+..+1/2^n=1/2×[1-(1/2)^n/(1-1/2)]=1-1/2^n
3、
b1=a2-a1/2
b2=a3-a2/2
....
bn=an-a(n-1)/2
b1+b2+...+bn=-a1/2+a2/2+a3/2+..+a(n-1)/2+an
Sn=a1/2+a2/2+...+an/2+an/2-a1
Sn=Tn/2+n^2/2^n-1
2Sn=Tn+n^2/2^(n-1)-2
Tn=2Sn-n^2/2^(n-1)+2
=2(1-1/2^n)-n^2/2^(n-1)+2
=4-(n^2+1)/2^(n-1)