设等比数列公比为k,第i项为a{i}；S{N}表前n项和于是S{N}=a{1}+k*a{1}+(k^2)*a{1}+……+[k^(k-1)]*a{1}kS{N}=k*a{1}+(k^2)*a{1}+……+[k^(k-1)]*a{1}+(k^k)*a{1}下式减上式,得(k-1)S{N}=a{1}*(k^k-1)当k不等于...