[X版高考数学人教A版·数学文全程复习方略配套课件5.3等比数列及其前n项和共53张PPT.pptVIP

【例2】设数列{an}的前n项和为Sn,已知a1=1,Sn+1=4an+2. (1)设bn=an+1-2an，证明数列{bn}是等比数列. (2)在(1)的条件下证明 是等差数列,并求an. 【解题指南】(1)利用Sn+1=4an+2，寻找bn与bn-1的关系. (2)先求bn，再证明数列 是等差数列，最后求an. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 【规范解答】(1)由a1=1,及Sn+1=4an+2, 有a1+a2=4a1+2,a2=3a1+2=5, ∴b1=a2-2a1=3. 由Sn+1=4an+2 ① 知当n≥2时，有Sn=4an-1+2 ② ①-②得an+1=4an-4an-1， ∴an+1-2an=2(an-2an-1) 又∵bn=an+1-2an，∴bn=2bn-1, ∴{bn}是首项b1=3,公比为2的等比数列. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. (2)由(1)可得bn=an+1-2an=3·2n-1, ∴数列 是首项为 公差为 的等差数列. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 【反思·感悟】在证明本题时，首先利用转化的思想，把 Sn+1=4an+2转化为an+1与an的关系，然后作商 或 在作商时，无论使用 还是 都要考虑比值中是否 包含了 这一项，这是很容易被忽视的地方. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 【变式训练】数列{an}的前n项和为Sn，若an+Sn=n,cn=an-1，求证:数列{cn}是等比数列. 【证明】∵an+Sn=n, ∴a1+S1=1,得a1= ∴c1=a1-1= 又an+1+Sn+1=n+1, ∴2an+1-an=1,即2(an+1-1)=an-1. 又∵a1-1= 即 ∴数列{cn}是以 为首项，以 为公比的等比数列. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 等比数列的性质及应用 【方法点睛】等比数列的常见性质 (1)若m+n=p+q=2k(m,n,p,q,k∈N*),则am·an=ap·aq=ak2; (2)通项公式的推广：an=am·qn-m(m,n∈N*); (3)若数列{an},{bn}(项数相同)是等比数列，则{λan}, (λ≠0)仍然是等比数列; Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. (4)在等比数列{an}中，等距离取出若干项也构成一个等比数列，即an,an+k,an+2k,an+3k,…为等比数列，公比为qk; (5)公比不为-1的等比数列{an}的前n项和为Sn,则Sn,S2n-Sn,S3n-S2n仍成等比数列，其公比为qn，当公比为-1时，Sn,S2n-Sn,S3n-S2n不一定构成等比数列. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 【例3】(1)已知各项均为正数的等比数列{an}中,a1·a2·a3 =5,a7·