Question

设{an}是首项为1的正项数列,且(n+1)a(n+1)^2-nan^2+ana(n+1)=0,(n∈N*),求它的通项公式(要迭代法）

Answer 1

(n+1)a(n+1)^2-nan^2+ana(n+1)=0
[(n+1)a(n+1)-nan] [ a(n+1) + an] =0
(n+1)a(n+1)-nan=0 or a(n+1)+an=0 (rejected)
a(n+1)/an = n/(n+1)
an/a(n-1) = (n-1)/n
an/a1 = 1/n
an= 1/n

追问

谢谢啊，答案对了，可是我不明白an/a(n-1) = (n-1)/n怎么变成an/a1 = 1/n的

追答

an/a(n-1) = (n-1)/n
[an/a(n-1)]. [a(n-1)/a(n-2)].....[a2/a1]
=[(n-1)/n].[(n-2)/(n-1)]....[1/2]
an/a1= 1/n
an = a1(1/n) =1/n

Answer 2

解：因式分解后，不能得(n+1)a(n+1)-nan=0 即an/a(n-1) = (n-1)/n
则an=a1*(a2/a1)*(a3/a2)*……*(an/a(n-1))
=1*1/2*2/3*3/4……*(n-1)/n
=1/n