已知一等差数列{an}的前n项和Sn=n^2-3n+1,求an

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$已知一等差数列{an}的前n项和Sn=n^2-3n+1,求an$

已知一等差数列{an}的前n项和Sn=n^2-3n+1,求an
已知一等差数列{an}的前n项和Sn=n^2-3n+1,求an

已知一等差数列{an}的前n项和Sn=n^2-3n+1,求an
首先要说的是,这道题肯定是错误的
因为等差数列的前N项和的表达式不可能含有常数项
如果Sn=n^2-3n,那么
an=Sn-S(n-1)
=n^2-3n-((n-1)^2-3(n-1)+1)
=n^2-3n-n^2+2n-1+3n-3
=2n-4

Sn=n^2-3n+1，
Sn-1=(n-1)^2-3(n-1)+1，
an=Sn-Sn-1
=n^2-3n+1-[(n-1)^2-3(n-1)+1]
=n^2-3n+1-n^2+2n-1+3n-3-1
=2n-4

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