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
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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
求a2的值
2.数列{an}的通项公式
![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](/uploads/image/z/2935279-55-9.jpg?t=%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5a1%3D1%2C2Sn%2Fn%3Da%28n%2B1%29-n%5E2%2F3-n-2%2F3+%2Cn%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0)
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
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
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