Question

已知数列{an}中,a1=1,an+1=(n/n+1)an,求an的通向公式,用叠加法

Answer 1

2a(n
1)=an
1(1)
2an=a(n-1)
1(2)
(1)-(2):
2[a(n
1)-an]=an-a(n-1)
[a(n
1)-an]/[an-a(n-1)]=1/2
a1=4,2a2=a1
1,a2=5/2,a3=7/4
(a2-a1)=-3/2*(1)
即数列{a(n
1)-an}是一个以-3/2为首项，公比为1/2的等比数列
a3-a2=-3/4*(2)
a4-a3=-3/8*(3)
....
an-a(n-1)*(n-1)
(1)
(2)
(3)
..(n-1):
an-a1=(-3/2)[1-(1/2)^(n-1)]/(1-1/2)
=3-3/2^(n-1)
an=7-3/2^(n-1)