1.在数列{an}中,a1=1,a(n+1)=(1+1/n)an+(n+1)/2的n次方

a(n+1)=[(n+1)/n]an+(n+1)/2^n
两边同除以n+1
a(n+1)/(n+1)-(an)/n=1/2^n
(an)/n-a(n-1)/(n-1)=1/2^(n-1)
.
(a2)/2-a1=1/2
叠加,中间项减去
a(n+1)/(n+1)-a1=1/2^n+1/2^(n-1)+...+1/2
=(1/2)(1-1/2^n)/(1-1/2)
=1-1/2^n
a(n+1)/(n+1)=2-1/2^n
an/n=2-1/2^(n-1)
(1)通项公式 bn=an/n=2-1/2^(n-1)
(2) an=2n-n/2^(n-1)
设前n项和为Sn=∑2n-∑n/2^(n-1)
Cn=∑2n Tn=∑n/2^(n-1)
则Cn=2[n(n+1)/2]=n(n+1)=n^2+n
Tn=1+2/2+3/2^2+.+n/2^(n-1)
(1/2)Tn=1/2+2/2^2+3/2^3+...+n/2^n
Tn-(1/2)Tn=1+1/2+1/2^2+.+1/2^(n-1)-n/2^n
=(1-1/2^n)/(1-1/2)-n/2^n
(1/2)Tn=2(1-1/2^n)-n/2^n
Tn=4-4/2^n-2n/2^n=4-(n+2)/2^(n-1)
所以Sn=n^2+n+4-(n+2)/2^(n-1)

1 a(n+1)=(n+1)an/n+(n+1)/2的n次方
可得bn=a-1/2的n次方
2 又a1=a-1/2,a=3/2
Sn=(1+2+……+n)a-(1+2+……+n)(1/2+……+1/2的n次方）
=3/4n(n+1)-1/2(n+1)n(1-2乘以1/2的n次方

2

两边除以(n+1)，然后迭代可以求得an/n的通项公式。

题目有没有错啊或者你没讲清楚