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已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)

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已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)
S3n=3na1+3n(3n-1)d
3(S2n-Sn)=3(2na1+2n(2n-1)d/2-na1-n(n-1)d/2)
=3na1+3n(3n-1)d
所以S3n=3(S2n-Sn)
再问: 3(S2n-Sn)=3(2na1+2n(2n-1)d/2-na1-n(n-1)d/2) 请问这一步怎么来的?为什么用除?
再答: S3n=3na1+3n(3n-1)d/2 3(S2n-Sn)=3(2na1+2n(2n-1)d/2-na1-n(n-1)d/2) =3(2na1+(4n^2-2n)d/2-na1-(n^2-n)d/2) =3(na1+(3n^2-n)d/2) =3na1+3n(3n-1)d/2 所以S3n=3(S2n-Sn) 3(S2n-Sn)=3(2na1+2n(2n-1)d/2-na1-n(n-1)d/2) 这里分别用求和公式求出 S3n S2n Sn 得到