已知数列{an}的前n项和为Sn,a1=1/2，且Sn=n^2An-n(n-1)

因Sn=n^2An-n(n-1)所以Sn-S(n-1)=An即An=n^2An-n(n-1)-[(n-1)^2A(n-1)^-(n-1)（n-2)]整理得 (n+1)An-(n-1)A(n-1)=2 An=2/(n+1)+{(n-1)/(n+1)}A(n-1) A1=1/2 A2=5/6 A3=11/12 1/2=1/(1*2) 5/6=5/(2*3) 11/12=11/(3*4)所以 An=[n(n+1)-1]/n(n+1)2) Sn=n^2An-n(n-1)=n^2[n(n+1)-1]/n(n+1)-n(n-1) 整理得 Sn=n^3/(n+1) 收起