a1=1 数列 （n+1)an+1^2-nan^2+an+1an=0 求通项公式

A(n+1)表示第n+1项(n+1)A(n+1)^2-nAn^2+A(n+1)An=0n(A(n+1)+An)(A(n+1)-An)+A(n+1)(A(n+1)+An)=0(A(n+1)+An)[nA(n+1)-nAn+A(n+1)]=0(A(n+1)+An)[(n+1)A(n+1)-nAn]=0因为{An}是首项为1的正项数列,因此A(n+1)+An大于0,...