《2022年数列求通项的方法总结-高三复习优质材料 .pdf》由会员分享，可在线阅读，更多相关《2022年数列求通项的方法总结-高三复习优质材料 .pdf（14页珍藏版）》请在taowenge.com淘文阁网|工程机械CAD图纸|机械工程制图|CAD装配图下载|SolidWorks_CaTia_CAD_UG_PROE_设计图分享下载上搜索。
1、学习必备欢迎下载数列求通项明师教育项目组望习才求通项方法总结11111.(1)()(1)2.:(2)(2)()3.(1)(2)()(3)()(4):1nnnnnnnnnnnnnSf 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：递推关系为构造新数列：递推关系为类型：111111212111()2fn13()4(,0)nnnnnnnnnnnnnnnnnnnnnapaqaxp axaapcapaqqq qqapaqaaxay axapaqap
2、q r hrrah待定系数的形式类型：（）两边同除以得类型类型：待定系数的形式类型：为常数 且分式型习题：一、观察法猜想通项公式,:(1)3,5,9,17,33(3)10,11,10,11,5714916(4)7,0,7,0,7(5)4,2,(6)1,2,3,424251017246810815 241 15 1329 61(7),(8)1,(9),3 15 35 63 995792 48 1632 64根据数列的前几项写出下面数列的一个通项公式19(4).8189189918999189991;n个，19:899919 109kkka个提示学习必备欢迎下载(2)1,11,111,1111提示
3、：111111119999(101)99kkkka个个二、根据递推关系求通项1.逐差法注意右边n 项和必须可以求和，否则此法行不通.)1(21245,)11(21124521,1,)1(21245)111211(2121)211(2121)(,2),211(21)2(11)1(1)(1)1(1:.),2(11,21.1111111221211nnnaaannnnnnkkkgaannnnnnngnaaannaaaannnknknnnnnnn的通项公式为数列时当时当记递推关系式可表示为解的通项公式求数列满足已知数列.),2(122,1,.1021nnnnnSnSSaaa求中备注：取倒数之后变成逐差
4、法。2.在数列 na中，已知,2,221111aaaannnnn，求该数列的通项公式.备注：取倒数之后变成逐差法。解：两边取倒数递推式化为:,211111nnnaa，即,211111nnnaa所以,2111212aa,2111323aa,2111434aa,nnnaa21111将以上1n个式子相加，得：,21212111321nnaa即,211211)211(2121212121132nnnna故.1222111nnnna文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编
5、码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C
6、2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I1
7、0F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS
8、2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R
9、3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP
10、1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I
11、2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1学习必备欢迎下载.,2,22.61111nnnnnnnaaaaaa求满足备注：倒数后变逐差法。2.逐商法.)2(;)1(*),(9log),2(3,1.43111nnnnnnnnnnbaNnaSnbnaaaa求求项和的前数列中，备注：逐商法的经典案例。.,0)1(,1.71221nnnnnnaaanaana求的正项数列是首项为备注：经典题，
12、通过简单变换就变成逐商法。n.,11,11:.,1,1342312,2,1,2)1(2)(2)(2:.),(2,1.31111111321321132111nnnnnnnnnnnnnnnnnnnnaanananaannanannnannaanannaaanaaaaaaaaanaaaaaanaaa得由本题还可这样做注时也满足该式当时当即解的通项公式求数列满足已知数列备注：经过简单变换之后是逐商法。.,)1()1()1()1(,2,.14444411nnnnnnnaaaaaaaa求中数列备注：取完对数之后变成了逐商法，也可以看作是等比数列。11111117.,*,1.2(1).(2)(2),(1,
13、2,1),1,.nkkkkkknkknakSSa akNaabknn nbknbbabn（2007陕西）各项不全为零的数列的前 项和为且求数列的通项公式对任意给定的正整数数列满足求数列的前 项和备注：一问奇偶项分析的典型案例。二问理科题，逐商法的应用。文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:C
14、N4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S1
15、0I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4
16、 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5
17、Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4
18、 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9
19、H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1
20、文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1学习必备欢迎下载3.类型一：1nnapaq11()nnnnapaqaxp ax，用待定系数法变成的形式。.,64*)(1,3,.211nnnnaaaNnnaa求有对于中备注：类型 1 经典题。.,*,.111111并求其通项是等比数列求证有对于它们满足数列项和的前nnnnnnnnnbaabnSaNnabbSna备注：经典题，通过an 和 sn的关系变换之后成为类型1。.3,)2(2,)2()21)(2(2),2(212,321,log,6loglog2,3:.,12,31,0.52)2(222111133136
21、216211nnnnnnnnnnnnnnnnnnnnnnabbbbbbbabaaaaanaaaaa从而继而可得则令解的通项公式求数列且满足数列备注：取完对数之后变成了类型一。.),4(21,1.1311nnnnaNnaaaa求备注：取完对数之后变成了类型一。4.类型二：1()(,0,1)nnapag nppp为常数 且111111()(),.,().nnnnnnnnnnnnaaag ng npbbbpppppg n将等式两边同除以则令则这样此种数列求通项的问题可转化为逐差法的问题当然这种数列的通项公式也常可用待定系数法解决 关键要根据选择适当的形式.,24,1.811nnnnnaaaaa求且的
22、首项备注：类型2 典型题。两种解法。.,log*111,2)2(;)1(*),2(22,2.152113221111的范围求都有对任意令求中，mmTNnbbbbbbTabaNnnaaaannnnnnnnnnnn备注：类型2 常见题型。两种解法。16.(20XX 年全国卷)在数列na中,14122333nnnSa,)(*Nn.求首项1a与通项na.文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1
23、文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:C
24、N4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S1
25、0I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4
26、 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5
27、Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4
28、 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9
29、H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1学习必备欢迎下载解：由题意得32231342111aSa，解得21a.又)22(3134341111nnnnnnnaaSSa，即1124nnnaa，设nnnnxaxa24211，利用待定系数法可得1x，又0421a，所以数列nna2是公比为4的等比数列.所以1442nnna，即nnna24.9.设数列na的前项为nS，已知2(1)nnnbabS(I)证明：当2b时，12nnan是等比数列；(II)求na的通项公式备注
30、：二问三种方法，一是除以n2，而是除以nb，或者待定系数法用逐差法来做。解析：由题意，在2(1)nnnbabS 中，令1n，得112(1)baba，12a由2(1)nnnbabS,得1112(1)nnnbabS(2,*)nnN,两式相减得：11()2(1)nnnnb aaba,即112nnnaba(2,*)nnN（）当2b时，由知，1122nnnaa于是11122(1)2nnnnanan212(1)2nnan(2,*)nnN又1 111 210a，所以12nnan是首项为1，公比为2 的等比数列（注：如果是求通项，也可化为等差数列来解决，解法如下：当2b时，由知，1122nnnaa，两边同时除
31、以2n得111222nnnnaa(2,*)nnN，即111222nnnnaa(2,*)nnN，2nna是等差数列，公差为12，首项为112a111(1)(1)222nnann,1(1)2nnan(易看出12nnan是等比数列,首项为 1,公比为 2)（）当2b时，由（）知，1122nnnan，即1(1)2nnan当2b时，由：112nnnaba，两边同时除以2n得11122 22nnnnaab，可设11()222nnnnaab，展开得11222 22nnnnaabb，与11122 22nnnnaab比较，得2122b，12b，1111()22222nnnnaabbb文档编码:CN4C2S10I
32、10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 H
33、S2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3
34、R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 Z
35、P1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7
36、I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档
37、编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4
38、C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1学习必备欢迎下载122nnab是等比数列，公比为2b，首项为11122bbb111()2222nnnabbbb，即111()2222nnnabbbb，11112(1)22()2222nnnnnbb
39、bbabbb.注：本问也可由待定系数法得到11112(2)22nnnnab abb进而求出通项.,322,:.2123,321221,322,3,322333,223,222,1)1()1(2,12:.,12,1,.61111111112121211求之错位相减法显然可用项和的前需求数列的过程中在用叠加法求注运用叠加法可求得则令得两边同除以即两式相减得则解的通项公式求数列且满足项和为的前数列nnbnanbnbbabnaanaanaaannSannSaannSaaSnannnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn.,134,2,.911nnnnanaaaa求中备注：类型二典型
40、题，也可以使用待定系数法。17.已知数列na满足21123451nnaanna，求数列na的通项公式.解：设221(1)(1)2()nnax ny nzaxnynz,212(2)(),nnaaxnyx nzxy即与21123451nnaanna，比较系数得:3324,10,518xxyxyzxyz解得，则31018xyz，故2213(1)10(1)182(31018)nnannann,则2123(1)10(1)18231018nnannann，故231018nann为以213 110 11813132a为首项，以 2 为公比的等比数列，因此2131018322nnann，即42231018nn
41、ann.文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档
42、编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4
43、C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I
44、10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 H
45、S2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3
46、R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 Z
47、P1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1学习必备欢迎下载5.类型三：11()(,)nnnapaqar np q为非零的常数),(.32123,21,23,)31(913),31(34,913,34,)
48、31(,31,1:.32123:.131;)31()31(913131,)(31),:(31,13134:.),2(3134,913,34.9221111111112121121请采用解法一答题如果高考中遇到这类解建议解得故可设解为由上知特征方程的两个解法二并整理得得由得由也可化为原递推关系式可化为这一过程省略掉如果是参加高考注得解方程解法一的通项公式求数列且满足数列nnnnnnnnnnnnnnnnnnnnnnnnnabababaaabaaaaaaaaaaaaaaaxxxxanaaaaaa.),3,2,1(,3235,35,1.161221nnnnnnSnnanaaaaaa项和的前求数列满足设
49、数列.274)323(5,272)323(52)12()23(85232)12()23(214522,2)12()23(214522),12(3452),12(345),12(334503533,3533,53,2,5)2(32),:(3,265:.,565,6,3.11121111111111111111111111122121221nnnnnnkknkknkkknnnnnnnnnnnnnnnnkknnnnnnnnnnnnnnnnnnnnnnankkaanaanaanbnkbbnbbnbbaabnaaaaxxxxanaaaaaa从而即则设原递推关系式可化为这一过程省略掉如果是参加高考注得解方
50、程解的通项公式求数列且满足数列备注：三项递推关系再加上一个f（n）6.类型四：1(,0)nnnpaqap q r hrrah为常数 且有如下结论:设方程pxqxrxh(此方程可在递推关系式中将),1得到都换成 xaann的两根为,(1)若1,a 且,则11nnnnaaprapra,即数列 nnaa等比数列;文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 HS2F5Z3R3G4 ZP1T9H7I2S1文档编码:CN4C2S10I10F4 H