怎样推出等差数列项的个数的奇偶性质:若共有2n项,S2n=n(an+a(n+1));S偶/S奇=a(n+1)/an;若共有2n+1项,S(2n+1)=(2n+1)*a(n+1);S偶—S奇=-a(n+1);S偶/S奇=n/(n+1).速求,急用

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