設(shè)原數(shù)列首項為a,公差為d,
原數(shù)列依次為a,a+d,a+2d,a+3d,.,a+2nd
奇數(shù)項為：a,a+2d,a+4d,.,a+2nd
奇數(shù)項和：S奇 = [a + (a+2nd)](n+1)/2 = (a+nd)(n+1)
偶數(shù)項為：a+d,a+3d,a+5d,.,a+(2n-1)d
偶數(shù)項和：S偶 = [(a+d) + (a+2nd-d)]n/2 = (a+nd)n
S奇/S偶 = (n+1)/n
說明：
本題只需用到等差數(shù)列求和公式：(首項+尾項)×項數(shù)÷2