等差数列{an},当项数分别为奇数或偶数时,求奇数项的和（S奇）跟偶数项的和（S偶）.
问题描述：
等差数列{an},当项数分别为奇数或偶数时,求奇数项的和（S奇）跟偶数项的和（S偶）.
要有详细的,便于理解的推导过程.
答
等差数列an,设公差为d,则an+1-an=d
对奇数项或偶数项,相邻两项中间间隔一项,则有an+2-an=2d
∴S奇=a1+a3+...+a(2k-1) (k=1,2,3...)
=(a1+a(2k-1))*k/2
=(a1+a1+(k-1)*2d)*k/2
=k*a1+k(k-1)d
=k*a1+k²d-kd
S偶=a2+a4+...+a(2k) (k=1,2,3...)
=(a2+a(2k))*k/2
=(a2+a2+(k-1)*2d)*k/2
=k*a2+k(k-1)d
=k*(a1+d)+k²d-kd
=k*a1+k²d