Question

已知数列{an}的前几项和为sn，且a1=2，nan+1=sn+n(n+1)/3.从{an}中抽出部分项

请明确一下结果。谢谢

Answer 1

na(n+1)=Sn+n(n+1)/3
Sn = na(n+1) -n(n+1)/3
an = Sn - S(n-1)
=na(n+1) -n(n+1)/3 - (n-1)an + (n-1)n/3
n( a(n+1) -an) =n(n+1)/3 -(n-1)n/3
= (2/3)n
a(n+1) -an = 2/3
=>{an}是等差数列, d=2/3
an -a1 = 2(n-1)/3
an =2(n+2)/3

追问

第二问的问题答案请再重复一下

追答

第二问?