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