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命题一 (3+5~(1/2))~n+(3-5~(1/2))~n能被2~n整除(n∈N) 这是中学数学中一道十份常见的题目,《数学教学通讯》1992年第4期吴跃生老师给出了命题一的推广,即命题二 [2((1+5~(1/2))/2)~(2k-1)]~n+[2((1-5~(1/2))/2)~(2k-1)]~n能被2~n整除(n∈N,k∈n) 而[2((1+5(1/2)/2))~(2k-1)]~n+[2((1-5(1/2)/2))~(2k-1)]~n=2~n[((1+5(1/2)/2))~(2~(k-1)n)+[2(((1-5(1/2)/2))~(2~(k-1)n],于是命题二等价于命题三 ((1+5(1/2)/2))~(2~(k-1)n)+((1-5(1/2)/2))~(2~(k-1)n)是整数(n∈N,k∈N),事实上,命题三可以进一步推广成
Proposition 1 (3+5~(1/2))~n+(3-5~(1/2))~n is divisible by 2~n (n∈N) This is a common topic in middle school mathematics , “Mathematics Teaching Newsletter” No. 4 of 1992, Professor Wu Yuesheng gave a generalization of Proposition 1, ie Proposition 2 [(2((1+5~(1/2))/2)~(2k-1)]~n+ [2((1-5~(1/2))/2)~(2k-1)]~n is divisible by 2~n(n∈N,k∈n) and [2((1+5( 1/2)/2))~(2k-1)]~n+[2((1-5(1/2)/2))~(2k-1)]~n=2~n[((1 +5(1/2)/2))~(2~(k-1)n)+[2(((1-5(1/2)/2))~(2~(k-1)n ], then proposition two is equivalent to proposition three ((1+5(1/2)/2))~(2~(k-1)n)+((1-5(1/2)/2)) ~(2~(k-1)n) is an integer (n∈N, k∈N). In fact, Proposition 3 can be further generalized