2022年中职数学基础模块下册《等差数列》word教案 .pdf

名师精编优秀教案等差 数 列教学目的：1.要求学生掌握等差数列的概念2.等差数列的通项公式，并能用来解决有关问题。教学重点：1.要证明数列 an 为等差数列，2.等差数列的通项公式：an=a1+(n-1)d(n 1,且 nN*).教学难点：等差数列“等差”的特点。公差是每一项（从第2 项起）与它的前一项的关绝对不能把被减数与减数弄颠倒。教学过程：一、引导观察数列：（1）1，3，5，7，9，11，（2）3，6，9，12，15，18，（3）1，1，1，1，1，1，1，（4）3，0，3，6，9，12，特点：从第二项起，每一项与它的前一项的差是常数“等差”二、得出等差数列的定义：一般地，如果一个数列从第2 项起，每一项与前一项的差等于同一个常数，那么这个数列叫做等差数列。注意：从第二项起，后一项减去前一项的差等于同一个常数。定义另叙述：在数列na 中，1nana=d（n N),d为常 数，则an 是等差数列，常数d 称为等差数列的公差。评注：1、一个数列，不从第2 项起，而是从第3 项起或第4 项起，每一项与它的前一项的差是同一个常数，此数列不是等差数列.如：（1）1，3，4，5，6，（2）1，0，12，14，16，18，20，2、公差 dR，当 d=0 时，数列为常数列；当d0 时，数列为递增数列；当d0 时，数列为递减数列。三、等差数列的通项公式：an=a1（n1）d 问题 1：已知等差数列an的首项为a1，公差为d，求daddadaadaddadaadaa3221134112312)()(由此归纳为dnaan)(11当1n时11aa（成立）dnaan)(11等差数列的通项公式四、应用例 1（1）求等差数列8，5，2，的第 20 项（2）401 是不是等差数列5，9，13，的项，如果是，是第几项？解：（1）由 a1=8,d=5 8=3,n=20,得:a20=8（201）（3）=49（2）由 a1=5,d=9（5）=4,得:an=5（n1）（4）即=4n1 由题意知，本题是要回答是否存在正整数n,使得若401=4 n 1 成立解这个关于n 的方程，得n=100 名师精编优秀教案即 401 是这个数列的第100 项例 2 在等差数列中，已知a5=10，a12=31，求首项a1与公差 d。解：由题意可知a1=2 解得：d=3 即这个等差数列的首项是2，公差是3。另解：由 an=ak（nk）d,知a12=a5（125）d,即 107d=31 解得 d=3 a5=a1（51）d 10=a143 解得 a1=2 即这个等差数列的首项是2，公差是3 作业：（1）求等差数列3，7，11，的第 4 项与第 10。（2）求等差数列10，8，6，的第 20 项。（3）100 是不是等差数列2，9，16，的项？如果是，是第几项？如果不是，说明理由。解：（1）由 a1=3，d=7 3=4 得a4=3+（41）4=15 a10=3+（101）4=39（2）由 a1=10，d=810=2，得 a20=10+（201）（2）=28（3）由 a1=2，d=92=7，得：=2+（n1）7=7n5 由题意知，7n 5=100 解得 n=15 即 100 是这个数列的第15 项。2.在等差数列an中，（1）已知 a410，a7=19，求 a1与 d；（2）已知 a3=9，a9=3，求 a12。解：（1）由题意知a1+3d=10 a1=1 a1+6d=19 d=3 即这个等差数列的首项为1，公差为3。（2）设等差数列 的首项为a1，公差为d，由题意可知：a1+（31）d=9 a1=11 a1+（91d）=3 d=1 这个数列的通项公式为an=12n a12=1212=0 另解：由 an=am+（nm）d，得 a9=a3+（9 3）d 3=9+（93）d d=1 a12=a3+（123）d=9+9（1）=0 五、小结：本节课首先要理解与掌握等差数列的定义及数学表达式a n+1 an=d(n N+)。其次，要会推导等差数列的通项公式：an=a1(n 1)d(n 1)，并掌握其基本应用。最后，还要注意一重要关系式：an=am(nm)d 的理解与应用。a111d=31 a14d=10 文档编码:CL7A4V8L10F3 HP10R3Z10W7M10 ZP4A1L6S10F7文档编码:CL7A4V8L10F3 HP10R3Z10W7M10 ZP4A1L6S10F7文档编码:CL7A4V8L10F3 HP10R3Z10W7M10 ZP4A1L6S10F7文档编码:CL7A4V8L10F3 HP10R3Z10W7M10 ZP4A1L6S10F7文档编码:CL7A4V8L10F3 HP10R3Z10W7M10 ZP4A1L6S10F7文档编码:CL7A4V8L10F3 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0 0九 年 十 一 月附件 2 优质课评选推荐表序号：作者姓名张海青性别男年龄37 职称中学一级工作单位义马市第二高级中学联系地址义马市第二高级中学邮编472300 课题等 差 数 列教学设计简要说明：等差数列是高中数学的一个重要知识点。本节学习了等差数列的定义以及通项公式。先通过 4 个例子引导学生分析,让学生观察特点，总结出等差数列的概念。进而引导学生推导等差数列的通项公式。有通项公式an=a1+(n1)d进一步变形得到an=ak+(nk)d。接下来通过例题加强对等差数列概念的理解，对通项公式以及变式掌握应用。之后通过课堂练习来反馈学生的学习情况。在课堂的最后环节通过学生作小节，来培养学生的自主学习，勇于探索的学习习惯。推荐意见：签字（盖章）：课例设计附后文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 ZP8Z5K3D9A3文档编码:CM2F9Z7S5K5 HI4Z10I8H6U10 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