题目不全,没有给出a1,原题应该有的吧.只好按你写的来解了：n≥2时,an=Sn-S(n-1)=n²·an-(n-1)²·a(n-1)(n²-1)an=(n-1)²·a(n-1)(n+1)(n-1)an=(n-1)²·a(n-1)(n+1)an=(n-1)a(n-1)a1=0时,{an...题目中是有a1的，a1=1.请您再按照这个条件算一下吧。谢谢。就说应该给出a1的值的。n≥2时，an=Sn-S(n-1)=n²·an-(n-1)²·a(n-1)(n²-1)an=(n-1)²·a(n-1)(n+1)(n-1)an=(n-1)²·a(n-1)(n+1)an=(n-1)a(n-1)an/a(n-1)=(n-1)/(n+1)a(n-1)/a(n-2)=(n-2)/n…………a2/a1=1/3连乘an/a1=(1/3)(2/4)....[(n-1)/(n+1)]=[1×2×...×(n-1)]/[3×4×...×(n+1)]=2/[n(n+1)]an=2a1/[n(n+1)]=2×1/[n(n+1)]=2/[n(n+1)]数列{an}的通项公式为an=2a1/[n(n+1)]