1.在数列{an}中,a1=1,a(n+1)=(1+1/n)an+(n+1)/2的n次方 (1)设bn=an/n,求数列{bn}的通项公式(2)求数列{an}的前n项和1 n和n+1均为a的下标
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![1.在数列{an}中,a1=1,a(n+1)=(1+1/n)an+(n+1)/2的n次方 (1)设bn=an/n,求数列{bn}的通项公式(2)求数列{an}的前n项和1 n和n+1均为a的下标](/uploads/image/z/12007574-62-4.jpg?t=1.%E5%9C%A8%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D1%2Ca%28n%2B1%29%3D%281%2B1%2Fn%29an%2B%28n%2B1%29%2F2%E7%9A%84n%E6%AC%A1%E6%96%B9+%EF%BC%881%EF%BC%89%E8%AE%BEbn%3Dan%2Fn%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C1+n%E5%92%8Cn%2B1%E5%9D%87%E4%B8%BAa%E7%9A%84%E4%B8%8B%E6%A0%87)
1.在数列{an}中,a1=1,a(n+1)=(1+1/n)an+(n+1)/2的n次方 (1)设bn=an/n,求数列{bn}的通项公式(2)求数列{an}的前n项和1 n和n+1均为a的下标
1.在数列{an}中,a1=1,a(n+1)=(1+1/n)an+(n+1)/2的n次方
(1)设bn=an/n,求数列{bn}的通项公式
(2)求数列{an}的前n项和
1 n和n+1均为a的下标
1.在数列{an}中,a1=1,a(n+1)=(1+1/n)an+(n+1)/2的n次方 (1)设bn=an/n,求数列{bn}的通项公式(2)求数列{an}的前n项和1 n和n+1均为a的下标
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)
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
2
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