设原数列首项为a,公差为d,
原数列依次为a,a+d,a+2d,a+3d,.,a+2nd
奇数项为：a,a+2d,a+4d,.,a+2nd
奇数项和：S奇 = [a + (a+2nd)](n+1)/2 = (a+nd)(n+1)
偶数项为：a+d,a+3d,a+5d,.,a+(2n-1)d
偶数项和：S偶 = [(a+d) + (a+2nd-d)]n/2 = (a+nd)n
S奇/S偶 = (n+1)/n
说明：
本题只需用到等差数列求和公式：(首项+尾项)×项数÷2