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第三届美国数学邀请赛试题中有这样一道题:选取一列整数a_1,a_2,a_3,…,使得每个n≥3都有a_n=a_(n-1)-a_(n-2),若该数列的前1492项之和等于1985,前1985项之和等于1492,那么前2001项之和是多少? 这是一道很好的数列题,它有多种解法,现介绍—种较为巧妙的解法。∵ a_n=a_(n-1)-a_(n-2) ∴ a_n=(a_(n-2)-a_(n-3))-a_(n-2)=-a_(n-3)。(1) 这表明数列中的第一项和第四项、第二项和第五项、第三项和第6项,……互为相反数重复使用(1)可得 a_n=-a_(n-3)=-(-a_(n-6))=a_(n-6)。(2) 这表明这个数列中的各项是以6为周期重复出现的。
The third session of the American Mathematical Invitational Exam has a question titled: Select a list of integers a_1, a_2, a_3, ... so that every n≥3 has a_n=a_(n-1)-a_(n-2). The sum of the first 1492 items of the series is equal to 1985, and the sum of the previous 1985 items is equal to 1492. What is the sum of the previous items of 2001? This is a good series of questions. It has many solutions, and introduces a more clever solution. . ∵ a_n=a_(n-1)-a_(n-2) ∴ a_n=(a_(n-2)-a_(n-3))-a_(n-2)=-a_(n-3). (1) This indicates that the first and fourth terms, the second and fifth terms, the third and the sixth terms of the sequence, ... are inversely numbered and used repeatedly (1) available a_n=-a_ ( N-3)=-(-a_(n-6))=a_(n-6). (2) This shows that the items in this series repeat with a period of 6 cycles.