题目

数列{An}中,A1=1,An大于0,（n+1）*An+1的平方-n*An的平方+An+1*An=0,（n属于N*）求An的通项公式

提问时间：2020-10-15

答案

(n+1)*(A[n+1])^2-n*(A[n])2+(A[n+1])*A[n]
=n((A[n+1])^2-(A[n])^2)+((A[n+1])^2+A[n+1]*A[n])
=n(A[n+1]+A[n])(A[n+1]-A[n])+A[n+1]*(A[n+1]+A[n])
=(A[n+1]+A[n])(nA[n+1]-nA[n]+A[n+1])
=(A[n+1]+A[n])((n+1)A[n+1]-nA[n])
=0
又因An>0
所以,只有 (n+1)A[n+1]-nA[n]=0
即 (n+1)A[n+1]=nA[n]
A[n+1]/A[n]=n/(n+1)
将上式从1取到n-1,然后两边相乘就有：
A[2]/A[1]=1/2
A[3]/A[2]=2/3
A[4]/A[3]=3/4
.
A[n]/A[n-1]=(n-1)/n
所以 A[n]=1/n*A1=1/n
即 A[n]=1/n