a1=1/2,an=(n²/n²-1)a[n-1]+n/(n+1) (n≥2) 则数列的通项an=

题目

a1=1/2,an=(n²/n²-1)a[n-1]+n/(n+1) (n≥2) 则数列的通项an=

答案

an=[n²/(n²-1)]a(n-1)+n/(n+1)
[(n+1)/n]an=[n/(n-1)]a(n-1)+1
设bn=[(n+1)/n]an,b1=2a1=1,
bn=b(n-1)+1
bn=b1+n-1=n
[(n+1)/n]an=bn=n
an=n^2/(n+1)