等差数列,当项数为2n+1,如何推导S奇－S偶=a1+nd

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)=a1+nd证明过程中a用大写了。希望能帮到你，不懂可以问