对于正整数集合A={a1,a2,…,an}(n∈N*,n≥3),如果去掉其中任意一元素ai(i=1,2,…,n)之后,剩余的所有元素组成的集合都能分为两个交集为空集的集合,且这两个集合的所有元素之和相等,就称集合A为“平衡集”.
(Ⅰ)判断集合Q={1,3,5,7,9}是否是“平衡集”并说明理由;
(Ⅱ)求证:若集合A是“平衡集”,则集合A中元素的奇偶性都相同;
(Ⅲ)证明:四元集合A={a1,a2,a3,a4},其中,a1<a2<a3<a4不可能是“平衡集”.
(Ⅰ)判断集合Q={1,3,5,7,9}是否是“平衡集”并说明理由;
(Ⅱ)求证:若集合A是“平衡集”,则集合A中元素的奇偶性都相同;
(Ⅲ)证明:四元集合A={a1,a2,a3,a4},其中,a1<a2<a3<a4不可能是“平衡集”.
更新时间:2021-10-24 09:55:47
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名校
【推荐2】已知
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【推荐1】已知集合
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【推荐2】已知集合
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