已知等差数列{An}的前n项和为Sn，公差为d.求证：
Sn,S2n-Sn,S3n-S2n,...,Skn-S(k-1)n(k>=3),成等差数列。
设等差数列的首先为a，公差为d，那么：
Sn=na+n(n-1)d/2
S2n=2na+2n(2n-1)d/2=2na+n(2n-1)d
S3n=3na+3n(3n-1)d/2
……
则：
(S2n-Sn)-Sn=S2n-2Sn=[2na+n(2n-1)d]-[2na+n(n-1)d]
=n(2n-1)d-n(n-1)d
=n^2*d
(S3n-S2n)-(S2n-Sn)=S3n-2S2n+Sn
=3na+3n(3n-1)d/2-2[2na+n(2n-1)d]+na+n(n-1)d/2
=n^2*d
……
所以，新数列是以Sn为首项，n^2*d为公差的等差数列。
答：设an=a+(n-1)d Sn=[a+a+(n-1)d]xn/2=[2a+(n-1)d]xn/2=18 an+an-1+an-2=a+(n-1)d+a+(n-2...详情>>
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