求证 等比数列中,Sm+n=Sn+qnSm=Sm+qmSn
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等比数列首项a1, 公比q，
Sm=a1(q^m-1)/(q-1), Sn=a1(q^n-1)/(q-1)，Sm+n=a1(q^(m+n)-1)/(q-1)
Sn+q^nSm=a1(q^n-1)/(q-1)+q^n*a1(q^m-1)/(q-1)
=[(q^n-1)+q^n*(q^m-1)]*a1/(q-1)
=[q^n-1+q^(n+m)-q^n)]*a1/(q-1)
=[q^(n+m)-1]*a1/(q-1)
=Sm+n
Sm+q^mSn=a1(q^m-1)/(q-1)+q^m*a1(q^n-1)/(q-1)
=[(q^m-1)+q^m*(q^n-1)]*a1/(q-1)
=[q^m-1+q^(n+m)-q^m)]*a1/(q-1)
=[q^(n+m)-1]*a1/(q-1)
=Sm+n
综上所述，
Sm+n=Sn+q^nSm=Sm+q^mSn
Sm=a1(q^m-1)/(q-1), Sn=a1(q^n-1)/(q-1)，Sm+n=a1(q^(m+n)-1)/(q-1)
Sn+q^nSm=a1(q^n-1)/(q-1)+q^n*a1(q^m-1)/(q-1)
=[(q^n-1)+q^n*(q^m-1)]*a1/(q-1)
=[q^n-1+q^(n+m)-q^n)]*a1/(q-1)
=[q^(n+m)-1]*a1/(q-1)
=Sm+n
Sm+q^mSn=a1(q^m-1)/(q-1)+q^m*a1(q^n-1)/(q-1)
=[(q^m-1)+q^m*(q^n-1)]*a1/(q-1)
=[q^m-1+q^(n+m)-q^m)]*a1/(q-1)
=[q^(n+m)-1]*a1/(q-1)
=Sm+n
综上所述，
Sm+n=Sn+q^nSm=Sm+q^mSn
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