高一数学[等差数列前n项与].ppt

2.3 等差数列的前n项和 高一数学必修五第二章 《数列》 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 复习巩固 1. an＝am＋(n－m)d 2.一般地，在等差数列{an}中， m＋n=p＋q a1＋an＝a2＋an－1＝a3＋an－2＝…. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例题讲解 例1 在等差数列{an}中，已知a1+a6=9, a4=7，求a3和a9. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例2 在等差数列{an}中，已知a1+a5=16, a2+a5=19 ，求数列{an}的通项公式. 例题讲解 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例3 在等差数列{an}中，已知 ， 且a1+a12=15,求数列{an}的通项公式. 例题讲解 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例4 已知四个数成等差数列，它们 的和为28，第二项与第三项之积为40， 求这四个数. 例题讲解 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 200多年前，高斯的算术老师提出 了下面的问题：1＋2＋3＋…＋100＝？ 据说高斯很快就算出了正确答案,你知 道他是如何计算的吗? (1＋100)＋(2＋99)＋…＋(50＋51)＝101×50＝5050. 提出问题 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 提出问题 倒序相加法: 凡是与首末两端等距离的两项之和 相等的数列，都可以用倒序相加法求前 n项和. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 等差数列前n项和公式: 知识传授 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例1 在等差数列{an}中，已知 ，求S7. 知识传授 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例2 已知一个等差数列{an}的前10项 的和是310，前20项的和是1220，求 这个等差数列的前n项和. 知识传授 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例3 2000年11月14日教育部下发了《关于在中小学实施