(1)若项数为偶数项2n则 s偶-s奇=nd s偶/s奇=An/An-1(n大于等于2)

1个回答

  • 应该有一个等差数列的条件,

    (1)S偶-S奇

    =[a2+a4+a6+.+a(2n)]-[a1+a3+a5+.+a(2n-1)]

    =(a2-a1)+(a4-a3)+(a6-a5)+.+[a(2n)-a(2n-1)]

    = d+d+d+.+d (n个d)

    =nd

    S偶=a2+a4+a6+.+a(2n)=[a(2)+a(2n)]*n/2=2a(n+1)*n/2=na(n+1)

    S奇=a1+a3+a5+.+a(2n-1)=[a(1)+a(2n-1)]*n/2=2a(n)*n/2=na(n)

    ∴ S偶/S奇=a(n+1)/a(n)

    你给的结果不对.

    (2)

    S奇-S偶

    =[a1+a3+a5+.+a(2n-1)+a(2n+1)]-[a2+a4+a6+.+a(2n)]

    =a1+(a3-a2)+(a5-a4)+(a7-a6)+.+[a(2n+1)-a(2n)]

    = a1+d+d+d+.+d (n个d)

    =a1+nd

    =a(n+1)

    S偶=a2+a4+a6+.+a(2n)=[a(2)+a(2n)]*n/2=2a(n+1)*n/2=na(n+1)

    S奇=a1+a3+a5+.+a(2n-1)+a(2n+1)=[a(1)+a(2n+1)]*(n+1)/2=2a(n+1)*(n+1)/2=(n+1)a(n+1)

    ∴ S偶/S奇=n/(n+1)