还有已知项数为2n+1等差数列,S奇-S偶=?S奇/S偶=?
问题描述：
还有已知项数为2n+1等差数列,S奇-S偶=?S奇/S偶=?
答
S奇-S偶=a（下脚标是n+1）
S奇/S偶=n+1：n
答
奇数项有n+1项,偶数项有n项
奇数项、偶数项分别成等差数列
S奇=(A1+A(2n+1))×(n+1)/2
=(A1+A1+2nd)×(n+1)/2
=(A1+nd)×(n+1)
=(n+1)A(n+1)
S偶=(A2+A(2n))×n/2
=(A1+d+A1+(2n-1)d)×n/2
=(A1+nd)×n
=nA(n+1)
S奇-S偶=(n+1)A(n+1)-nA(n+1)=A(n+1)
S奇/S偶=(n+1)A(n+1)/nA(n+1)=(n+1)/n