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高中《代数》第三册p81第18题是;“求证:1!2!3!…n!=(n!)~(n-1)/3·4~2·5~3…n~(n-2)”。此题较抽象,很多同学在解此题时无从下手,叙述不清。教学中,为了帮助学生探求解题途径,根据待证的恒等式的特点设计了一个“数表”:即将数1,2,3,…n填入(n-1)行n列表格中(如表一),然后用一直线将表格分为两部分,让学生观察,思考,不难发现数表中的所有数之积为(n!)~(n-1),直线右边部分所有数之积为3·4~2·5~3·…n~(n-2),直线左边部分所有数之积为1!·2!·3!…n!,其等式显然成立,学生对这种巧列数表的解题方法相当满意。
The 18th problem of p81 in the third book of high school algebra is: “Proof:1!2!3!...n!=(n!)~(n-1)/3·4~2·5~3...n~( N-2)”. This question is more abstract, many students can’t explain the problem when they solve the problem. In teaching, in order to help students explore ways to solve problems, according to the characteristics of the identity of the identity to be designed a “number table”: the number 1,2,3,... n is filled into (n-1) row n list (such as Table 1), then use a straight line to divide the table into two parts so that the students can observe and think. It is not difficult to find that the product of all the numbers in the table is (n!)~(n-1), and the number of the right part of the line is The product is 3·4~2·5~3·...n~(n-2) and the product of all the numbers on the left part of the line is 1!·2!·3!...n!, the equation is obviously established, and the students are The problem solving method of the sort of clever column number table is quite satisfactory.