Question

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

Answer 1

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