题目

数列an的前n项和为Sn,已知a1=1,2Sn/n=a(n+1)-n^2/3-n-2/3 ,n属于正整数

求a2的值

2.数列｛an｝的通项公式
 

提问时间：2020-08-02

答案

2Sn/n=a(n+1)-n^2/3 -n -2/3
6Sn = 3n(S(n+1) -Sn) - n^3 - 3n^2 - 2n
3nS(n+1) = 3(n+2)Sn +n^3 +3n^2 +2n
=3(n+2)Sn +n(n+1)(n+2)
3S(n+1)/[(n+1)(n+2)] = 3Sn/[n(n+1)] +1
S(n+1)/[(n+1)(n+2)] -Sn/[n(n+1)] =1/3
Sn/[n(n+1)] - S1/[1(1+1)] = (n-1)/3
Sn/[n(n+1)] = (2n+1)/6
Sn = (2n+1)n(n+1)/6
an = Sn -S(n-1)
= (1/6)[(2n+1)n(n+1) - (2n-1)(n-1)n ]
= n^2
a2 = 2^2= 4