设{an}是首项为1的正项数列,且（n+1）*[a(n+1)]^2-n*(an)^2+a(n+1)an=0(n=1,2,3,…),则它的通项公式an等于什么?注：其中an,a(n+1)中的n,（n+1）均为a的下标

Answer 1

n+1)*a(n+1)下标^2-n*an^2+a(n+1)下标*an=0
2*（n+1)*a(n+1)下标^2 - 2*n*an^2 + 2*a(n+1)下标*an=0
(2n+1)*a(n+1)下标^2-(2n+1)*an^2
+ a(n+1)下标^2 + an^2 + 2*a(n+1)下标*an=0
(2n+1)[a(n+1)下标^2-an^2] + [a(n+1)下标^2 + an^2]^2=0
(2n+1)[a(n+1)下标 -an] + a(n+1)下标^2 + an^2 =0
a(n+1)下标
=n/(n+1) *an
=n(n-1)/[(n+1)*n] *a(n-1)下标
=n*(n-1)*.*1 / [(n+1)*n*.*2] *a1
=1/(n+1)
an=1/n