已知{an}是等差数列 求证Sn，S2n-Sn，S3n-S2n成等差数列
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Sn=a1n+n(n-1)d/2
S2n=2a1n+n(2n-1)d
S3n=3a1n+3n(3n-1)d/2
S2n-Sn-Sn=2a1n+n(2n-1)d-2a1n+n(n-1)d=n^2d
S3n-S2n-(S2n-Sn)=S3n-2S2n+Sn=3a1n+3n(3n-1)d/2-4a1n-2n(2n-1)d+a1n+n(n-1)d/2
=n^2d
所以,Sn,S2n-Sn,S3n-S2n成等差数列
S2n=2a1n+n(2n-1)d
S3n=3a1n+3n(3n-1)d/2
S2n-Sn-Sn=2a1n+n(2n-1)d-2a1n+n(n-1)d=n^2d
S3n-S2n-(S2n-Sn)=S3n-2S2n+Sn=3a1n+3n(3n-1)d/2-4a1n-2n(2n-1)d+a1n+n(n-1)d/2
=n^2d
所以,Sn,S2n-Sn,S3n-S2n成等差数列
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