高三文科数学[数列[2.ppt

Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 灵活运用等差、等比数列 的公式与性质 类型一 根据数列通项公式、求和公式， 列方程组解决问题. 类型二 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 类型三 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. (1)设等比数列{an}的公比为q,前n项和为Sn,若Sn+1, Sn, Sn +2成等差数列，则q的值为 . -2 例1 = . 3 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例2 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 1.公式法 常用的公式有： (1)等差数列{an}的前n项和； (2)等比数列{an}的前n项和 (3)12+22+32+…+n2= . 类型四 数列求和 常用求和公式： Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2.倒序相加法 将一个数列倒过来排序，它与原数列相加时，若有公因式可提，并且剩余的项易于求和，则这样的数列可用倒序相加法求和. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 3.分组转化法 分析通项虽不是等差或等比数列，但它是等差数列和等比数列的和的形式,则可进行拆分，分别利用基本数列的求和公式求和，如求{n(n+1)}前n项的和. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 4.错位相减法 利用等比数列求和公式的推导方法求解， 一般可解决型如一个等差数列和一个等比数列对应项相乘所得数列的求和，如求数列{n·3n}的前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. 5.裂项相消法 把数列和式中的各项分别裂开后，消去一部分从而计算和的方法，它适用于通项为 的前n项求和问题。 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 求和： (1)Sn=(2-3×5)+(4-3×52)+…+(2n-3×5n); (2) 例3 (3)Tn=1×2+2×22+3×23+…+n×2n. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 若{an}成等差数列，{bn}成等比数列，则若求数列{an bn}的前n项和Sn，用错位相减法；若求数列{