Question

等差数列S奇：S偶的规律是怎么得出的?

Answer 1

∵等差数列,∴a(2n-1)-a(2n-3)=a(2n)-a(2n-2)=2d

S奇=a1+a3+...+a(2n-1)=[a1+(n-1)*2d]*n/2

S偶=a2+a4+...+a(2n)=[a2+(n-1)*2d]*n/2

∴S奇：S偶={[a1+(n-1)*2d]*n/2}：{[a2+(n-1)*2d]*n/2}

=[a1+(n-1)*2d]：[a2+(n-1)*2d]

=[a1+(n-1)*2d]：[a1+d+(n-1)*2d]

=1-d/[a1+d+(n-1)*2d]