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

（n+1)a(n+1)^2-nan^2+ana(n+1)=0
[(n+1)a(n+1)-nan][a(n+1)+an]=0
因为：an>0
所以,(n+1)a(n+1)-nan=0
a(n+1)=an*n/(n+1)
an=a(n-1)*(n-1)/n
a(n-1)=a(n-2)*(n-2)/(n-1)
...
a2=a1*1/2
所以,an=a1*1/2*2/3*...*(n-2)/(n-1)*(n-1)/n
=a1/n
=1/n