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问题试用数学归纳法证明(3+5~(1/2))~n+(3-5~(1/2)))~n能被2~n整除,其中n为任意自然数。这是一位刚学“数学归纳法”的高二学生提出的,她百思不得其解。我也未见过本题。既然是初学者提出,想不会太难,于是便从常规法入手。设a_n=(3+5~(1/2))~n+(3-5~(1/2))~n,则a_1=(3+5~(1/2))+(3-5~(1/2))=6,能被2整除。这说明“归纳基础”已具备。接下去只需在“归纳假设”; a_k能被2~k整除的基础上去证明a_(k+1)能被2~(k+1)整除,以完成数学归纳法的第二步。我的思路从a_(k+1)中析出a_k,目的是便于运用
Probability of the trial using mathematical induction (3 +5 ~ (1/2)) ~ n + (3-5 ~ (1/2))) ~ n divisible by 2 ~ n, where n is an arbitrary natural number. This is a sophomore who just learned “mathematical induction” proposed by her, puzzled. I have not seen this problem Since it is a beginner put forward, would like not too difficult, so they start from the conventional law. Let a_n = (3 + 5 ~ (1/2)) ~ n + (3-5 ~ (1/2)) ~ n, (1/2)) = 6, divisible by 2. This shows that “foundation of induction” already exists. In the following, only the “inductive hypothesis” is needed; a_k can be divided by 2 ~ k to prove that a_ (k + 1) can be divisible by 2 ~ (k + 1) to complete the second step of mathematical induction. My idea a_k from a_ (k + 1), the purpose is to facilitate the use