在数列｛an｝中，设S1=a1+a2+……an,s2=a(n+1)....

(1)S2-S1=[a(n+1)-a1]+[a(n+2)-a2]……+[a2n-an]=nd+nd+...+nd=n^2*d
类似得S3-S2=n^2*d。得证

(2)把a(n+1)=a1*q^n,a(n+2)=a2*q^n,...,a2n=an*q^n相加得
S2=(a1+a2+...+an)*q^n=S1*q^n
同理S3=S2*q^n
所以S1,S2,S3成等比数列

｛an｝是等差数列
设首项是a，公差是d
则S1=na+d*n(n-1)/2
a1+a2+……+a2n=2na+d*2n(2n-1)/2
a1+a2+……+a3n=3na+d*3n(3n-1)/2
所以S2=(a1+a2+……+a2n)-(a1+a2+……+an)=2na+d*2n(2n-1)/2-[na+d*n(n-1)/2]
=na+d/2*(4n^2-2n-n^2+n)
=na+d/2*(3n^2-n)

S3=(a1+a2+……+a3n)-(a1+a2+……+a2n)=3na+d*3n(3n-1)/2-[2na+d*2n(2n-1)/2]
=na+d/2*(9n^2-3n-4n^2+2n)
=na+d/2*(5n^2-n)

S1+S3=2na+d/2*(5n^2-n+n^2-n)=2na+d/2(6n^2-2n)
2S2=2*[na+d/2*(3n^2-n)]=2na+d/2(6n^2-2n)=S1+S3
所以S1,S2,S3也是等差数列

｛an｝是等比数列
S1=a(q^n-1)/(q-1)
S2=a(q^2n-1)/(q-1)-a(q^n-1)/(q-1)
S3=a(q^3n-1)/(q-1)-a(q^2n-1)/(q-1)
a/(q-1)是共同的，以后先不考虑
S1*S3=(q^n-1)[(q^3n-1)-(q^2n-1)]
=(q^n-1)(q^3n-q^2n)
=q^2n*(q^n-1)^2
S2^2=[(q^2n-1)-(q^n-1)]^2=(q^2