Question

证明.项数为奇数2n-1的等差数列｛an｝,有 S奇－S偶＝an,s奇／S偶＝n/n-1.

Answer 1

证明：由题意令此数列公差为d,则:a(n+1)-an=d,即an-a(n+1)=d又由通项公式得：a(2n-1)=a1+(2n-2)d=an+(n-1)dS奇－S偶＝(a1-a2)+(a3-a4)+...+(a(2n-3)-a(2n-2))+a(2n-1)=(n-1)*(-d)+an+(n-1)d=an求前2n-1项和得：S(2n-...