2022年数列复习总结求通项求和 .pdf
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1、数列求通项方法总结11111.(1)()(1)2.:(2)(2)()3.(1)(2)()(3)()4(5):nnnnnnnnnnnnnSf naSnSaaSSnSf aaaf naaf n观察猜想，用数学归纳法证明得由求方法；类型：得递推关系由递推关系求通项公式法：运用等差、等比数列定义与通项公式；累加（逐差）法：递推关系为累乘（逐商）法：递推关系为倒数法构造新数列：递推关系为1111111121211111()2112()3()4lnlnnnnnnnnnnnnnnnnnnnnnnnnknnnnapaqaxp axaapcapac qqqq qqaqp aqapaqaaxay axaapapa
2、pkap类型：待定系数的形式类型：方法：两边同除以得类型方法：待定系数类型：待定系数的形式类型：取对数成等比一、由递推关系求通项：构造新数列1、11211,.2nnnaaaann已知求2、112,.31nnnnaaaan已知求3、113,23,nnnaaaa已知求4、113,.521nnnnaaaaa已知求5、11111,.1nnnnsasas已知求6、1121,nnnanaaan已知求7、1113,nnnnnaaaaa数列满足=1,求8、111,32,.nnnnaaaa已知求9、12215521,333nnnnnaaaaaaa数列满足求、10、112,21.nnnnnaaaaaa为正项数列，
3、求数列求和一、错位相减法要点：（1）适用：an等差，bn等比，求 an?bn的和；（2）注意：乘公比，错位对齐；（3）讨论：公比q 是否为 1.32:.132nnnxxxxS求和文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文
4、档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT
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10、U2Y10.)Rx()2()1(.122,)(03.23211项和公式前，求数列令的通项公式；求数列且为等差数列北京nbxabaaaaaannnnnn二、裂项法要点：（1）常用的裂项方法；（2）注意：根据前后对称的特点，注意抵消后保留哪些项裂项相消法求和：把数列的通项拆成两项之差、正负相消剩下首尾若干项。常见拆项：111)1(1nnnn)121121(21)12)(12(1nnnn)211(21)2(1nnnn)2)(1(1)1(121)2)(1(1nnnnnnn)(1.1knn)12)(12(1.2nnnkn1.3)2)(1(1.4nnn1.111111(1),;(2),1 3 35 571
11、 23 234 345nLL求下面数列的前项和：.)12)(12()2(534312.2222nnnSn求和.,10,11.3nnnnan求项和为若前文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P10 HY3O4A5J4R5 ZH6A7Q3U2Y10文档编码:CT7R5I6K5P
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18、),(9log),2(3,143111nnnnnnnnnnbaNnaSnbnaaaa求求项和的前数列中，、证明等比112211.,:0,1,.2(1)(2).nnnnnnnnnnaaabaaabaabb数列满足下列条件求证：是等比数列；求的通项公式112.1,21.(1)1;(2).nnnnnaaaaaa数列满足求证：数列是等比数列求数列的通项公式.)2(b)1(,21),2(44,4.3n11nnnnnnaabnaaaa求是等差数列；求证：数列令满足数列.42S)1(),3,2,1(2,1)04.(4111nnnnnaSnnSnnaa）是等比数列；（数列证明：全国文档编码:CT7R5I6K5
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