2022年数学归纳法教案2 .pdf
名师精编精品教案数学归纳法授课人：康正班级：高二（2）班一、教材分析“数学归纳法”既是高中代数中的一个重点和难点内容，也是一种重要的数学方法。它贯通了高中代数的几大知识点：不等式，数列，三角函数在教学过程中，教师应着力解决的内容是：使学生理解数学归纳法的实质，掌握数学归纳法的证题步骤（特别要注意递推步骤中归纳假设的运用和恒等变换的运用）。只有真正了解了数学归纳法的实质，掌握了证题步骤，学生才能信之不疑，才能用它灵活证明相关问题。本节课是数学归纳法的第一节课，有两大难点：使学生理解数学归纳法证题的有效性；递推步骤中归纳假设的利用。不突破以上难点，学生往往会怀疑数学归纳法的可靠性，或者只是形式上的模仿而不知其所以然。这会对以后的学习造成极大的阻碍。根据本节课的教学内容和学生实际水平，本节课采用“引导发现法”和“讲练结合法”。通过课件的动画模拟展示，引发和开启学生的探究热情，通过“师生”和“生生”的交流合作，掌握概念的深层实质。二、教学目标1、知识和技能目标(1)了解数学推理的常用方法（归纳法）(2)了解数学归纳法的原理及使用范围。(3)初步掌握数学归纳法证题的两个步骤和一个结论。2、过程与方法目标通过对归纳法的复习，说明不完全归纳法的弊端，通过多米诺骨牌实验引出数学归纳法的原理，使学生理解理论与实际的辨证关系。在学习中培养学生探索发现问题、提出问题的意识，解决问题和数学交流的能力,学会用总结、归纳、演绎类比探求新知识。3.情感态度价值观目标通过对问题的探究活动，亲历知识的构建过程，领悟其中所蕴涵的数学思想；体验探索中挫折的艰辛和成功的快乐，感悟“数学美”，激发学习热情，培养他们手脑并用，多思勤练的好习惯和勇于探索的治学精神。名师精编精品教案三、教学重难点重点：（1）使学生理解数学归纳法的实质。（2）掌握数学归纳法证题步骤,尤其是递推步骤中归纳假设和恒等变换的运用。难点：数学归纳法的原理；四、教学方法：讲授法、引导发现法、类比探究法、多媒体辅助教学五、教学过程复习引入问题（1）袋中有 5 个小球，如何证明它们都是红色的？（完全归纳法）问题（2）某人站在学校门口，看到连续有20个男生进入学校，于是深有感触的说这个学校的学生都是男生。（不完全归纳法）新课讲解1、多米诺骨牌实验要使所有的多米诺骨牌一一倒下？需要几个步骤才能做到？（1）第一张牌被推倒（奠基作用）（2）任意一张牌倒下必须保证它的下一张牌倒下（递推作用）于是可以获得结论：多米诺骨牌会全部倒下。2、数学归纳法类比多米诺骨牌可得数学归纳法如下(1)验证当 n取第一个值0n（如0n=1 或 2 时）命题正确。(2)假设当nk时0(,)kN kn命题正确，证明1nk时命题也正确，这就是数学归纳法文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9名师精编精品教案经典例题讲解例 1：如果是等差数列，已知首项为a1，公差为 d，那an=a1+（n-1）d 对一切正整数都成立，试用数学归纳法证明证明：（1）当 n=1时，左边=a1，右边=a1等式成立（2）假设等 n=k 时成立，就是,)1(1dkaak那么daakk1ddka)1(1dka 1)1(1就是说当 n=k+1 时也成立，由（1）和（2），可知等式对任何正整数都成立例 2：用数学归纳法证明Nn，证明（1）当 n=1时，左边=1，右边=1，等式成立。（2）当 n=k 时，有2)12(.531kk则，当 n=k+1 时即 n=k+1 时，命题成立根据问可知，对nN，等式成立。上如证明对吗？为什么？正确解法证明：证明（1）当 n=1时，左边=1，右边=1，等式成立。（2）假设当 n=k 时，有2)12(.531kk则，当 n=k+1 时=k2+2(k+1)1na2)12(.531nn2 12(1)1(1)2(1)kkk135.(21)2(1)1kk135.(21)2(1)1 kk文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9名师精编精品教案=k22k1=（k+1）2 即当 n=k+1 时等式也成立。根据（1）和（2）可知，等式对任何nN 都成立。即 n=k+1 时，命题成立根据问可知，对nN，等式成立。例 3 用数学归纳法证明基础反馈（1）用数学归纳法证明：Nnaaaaaann,1111212在验证 n=1成立时，左边计算所得的结果是（）A1 B.a1C21aaD.321aaa（2）求证:(n+1)(n+2)(n+n)=2n?1?3?(2n-1)课堂小结（1）理解数学归纳法的原理（2）数学归纳法的两个步骤缺一不可，前者是基础，后者是递推依据，最终给出结论。（3）数学归纳法主要应用于解决与正整数有关的数学问题。作业:1.求证：1+2+3+n=21n(n+1)2.求证：1 22334 n(n1)六、板书设计：利用多媒体和黑板合理搭配七、课后反思1(1)(2)3nnn6)12)(1(3212222nnnn文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9文档编码:CD8A3K9C3D4 HS6V5K7O5P10 ZL4K2K10G2Z9
