...A={a1,a2,……an},其中ai∈R(1≤i≤n,n>2),L(A)表示和ai+aj(1≤...

发布网友 发布时间:2024-10-27 21:01

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热心网友 时间:2024-10-27 22:33

1)根据题中的定义可知:由2+4=6,2+6=8,2+8=10,4+6=10,4+8=12,6+8=14,得l(P)=5.
由2+4=6,2+8=10,2+16=18,4+8=12,4+16=20,8+16=24,得l(Q)=6
2)证明:因为ai+aj(1≤i<j≤n)最多有C2n=
n(n-1)2个值,所以l(A)≤
n(n-1)2.
又集合A=2,4,8,,2n,任取ai+aj,ak+al(1≤i<j≤n,1≤k<l≤n),
当j≠l时,不妨设j<l,则ai+aj<2aj=2j+1≤al<ak+al,
即ai+aj≠ak+al.当j=l,i≠k时,ai+aj≠ak+al.
因此,当且仅当i=k,j=l时,ai+aj=ak+al.
即所有ai+aj(1≤i<j≤n)的值两两不同,
所以l(A)=
n(n-1)2

热心网友 时间:2024-10-27 22:31

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热心网友 时间:2024-10-27 22:32

(1) L(p)=5 L(Q)=6
(2)设m n p q 为互不相同数 如果2^m+2^n =2^p+2^q 2^m(1+2^(n-m)-2^(q-m)-2^(p-m))=0
1=2^(q-m)+2^(p-m)-2^(n-m) 不符合

热心网友 时间:2024-10-27 22:33

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