已知数列{an}满足：a1=1,an+1=(1+1/n)an+1/n(n∈N*) ①设bn=an/n,求数列{bn}的通项公式

Question

急求~~~~~~~~~~~

Answer 1

a(n+1)=(1+1/n)an+1/n
a(n+1)=(n+1)an/n+1/n
na(n+1)=(n+1)an+1
a(n+1)/(n+1)=an/n+1/n(n+1)
a(n+1)/(n+1)-an/n=1/n(n+1)

a(n+1)/(n+1)-an/n=1/n(n+1)
..........
a3/3-a2/2=1/2*3
a2/2-a1/1=1/1*2
以上等式相加得
a(n+1)/(n+1)-a1/1=1/1*2+1/2*3+.........+1/n(n+1)
a(n+1)/(n+1)-a1=1-1/2+1/2-1/3+.......+1/n-1/(n+1)
a(n+1)/(n+1)-a1=1-1/(n+1)
a(n+1)/(n+1)-a1=n/(n+1)
a(n+1)/(n+1)-1=n/(n+1)
a(n+1)/(n+1)=n/(n+1)+1
a(n+1)/(n+1)=(n+n+1)/(n+1)
a(n+1)/(n+1)=(2n+1)/(n+1)
a(n+1)/(n+1)=[(2n+2)-1]/(n+1)
a(n+1)/(n+1)=[2(n+1)-1]/(n+1)
所以an/n=(2n-1)/n
即bn=an/n=(2n-1)/n

Answer 2

na(n+1)=(n+1)an+1
所以a(n+1)/(n+1)=an/n+1/[n(n+1)]
即b(n+1)=bn+1/[n(n+1)]
累加法：bn=b1+(b2-b1)+(b3-b2)+…+(bn-b(n-1))
=1+1/(1*2)+1/(2*3)+…+1/[(n-1)n]
=1+(1-1/2)+(1/2-1/3)+…+(1/(n-1)-1/n)
=2-1/n本回答被提问者采纳