2022年数列题型及解题方法归纳总结90474 .pdf

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1、文德教育1 知识框架111111(2)(2)(1)(1)()22()nnnnnnmpqnnnnaq naaa qaad naandnn nSaanadaaaamnpq两个基等比数列的定义本数列等比数列的通项公式等比数列数列数列的分类数列数列的通项公式函数角度理解的概念数列的递推关系等差数列的定义等差数列的通项公式等差数列等差数列的求和公式等差数列的性质1111(1)(1)11(1)()nnnnmpqaa qaqqqqSnaqa aa amnpq等比数列的求和公式等比数列的性质公式法分组求和错位相减求和数列裂项求和求和倒序相加求和累加累积归纳猜想证明分期付款数列的应用其他掌握了数列的基本知识，特

2、别是等差、等比数列的定义、通项公式、求和公式及性质，掌握了典型题型的解法和数学思想法的应用，就有可能在高考中顺利地解决数列问题。一、典型题的技巧解法1、求通项公式（1）观察法。（2）由递推公式求通项。对于由递推公式所确定的数列的求解，通常可通过对递推公式的变换转化成等差数列或等比数列问题。(1)递推式为 an+1=an+d 及 an+1=qan（d，q 为常数）例 1、已知 an满足 an+1=an+2，而且 a1=1。求 an。例 1、解an+1-an=2为常数an是首项为1，公差为2 的等差数列an=1+2（n-1）即 an=2n-1 例 2、已知na满足112nnaa，而12a，求na=

3、？（2）递推式为 an+1=an+f（n）例 3、已知 na中112a，12141nnaan，求na.解：由已知可知)12)(12(11nnaann)121121(21nn令 n=1，2，（n-1），代入得（n-1）个等式累加，即（a2-a1）+（a3-a2）+（an-an-1）文德教育2 2434)1211(211nnnaan说明只要和f（1）+f（2）+f（n-1）是可求的，就可以由an+1=an+f（n）以 n=1，2，（n-1）代入，可得n-1 个等式累加而求an。(3)递推式为 an+1=pan+q（p，q 为常数）例 4、na中，11a，对于 n1（nN）有132nnaa，求na.

4、解法一：由已知递推式得an+1=3an+2，an=3an-1+2。两式相减：an+1-an=3（an-an-1）因此数列 an+1-an是公比为3 的等比数列，其首项为a2-a1=（31+2）-1=4 an+1-an=43n-1 an+1=3an+2 3an+2-an=43n-1 即 an=23n-1-1 解法二：上法得 an+1-an是公比为3 的等比数列，于是有：a2-a1=4，a3-a2=4 3，a4-a3=432，an-an-1=4 3n-2，把n-1个等式累加得：an=23n-1-1(4)递推式为 an+1=p an+q n（p，q 为常数）)(3211nnnnbbbb由 上 题 的

5、 解 法，得：nnb)32(23nnnnnba)31(2)21(32(5)递推式为21nnnapaqa思路：设21nnnapaqa,可以变形为：211()nnnnaaaa，想于是 an+1-an 是公比为 的等比数列，就转化为前面的类型。求na。文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5

6、HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I

7、3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5

8、HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I

9、3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5

10、HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I

11、3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文德教育3(6)递推式为 Sn与 an的关系式关系；（2）试用 n 表示 an。)2121()(1211nnnnn

12、naaSS11121nnnnaaannnaa21211上式两边同乘以2n+1得 2n+1an+1=2nan+2 则2nan 是公差为2 的等差数列。2nan=2+（n-1）2=2n 数列求和的常用方法：1、拆项分组法：即把每一项拆成几项，重新组合分成几组，转化为特殊数列求和。2、错项相减法：适用于差比数列（如果na等差，nb等比，那么nna b叫做差比数列）即把每一项都乘以nb的公比q，向后错一项，再对应同次项相减，转化为等比数列求和。3、裂项相消法：即把每一项都拆成正负两项，使其正负抵消，只余有限几项，可求和。适用于数列11nnaa和11nnaa（其中na等差）可裂项为：111111()nn

13、nnaadaa，1111()nnnnaadaa等差数列前 n项和的最值问题：1、若等差数列na的首项10a，公差0d，则前n项和nS有最大值。文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4

14、ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J

15、7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4

16、ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J

17、7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4

18、ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J

19、7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文德教育4（）若已知通项na，则nS最大100nnaa；（）若已知2nSpnqn，则当n取最靠近2qp的非零自然数时nS最大；2、若等差数列na的首项10a，公差0d，则前n项和nS有最小值（）若已知通项na，则nS最

20、小100nnaa；（）若已知2nSpnqn，则当n取最靠近2qp的非零自然数时nS最小；数列通项的求法：公式法：等差数列通项公式；等比数列通项公式。已 知nS（即12()naaaf n）求na，用 作 差 法：11,(1),(2)nnnSnaSSn。已知12()na aaf n求na，用作商法：(1),(1)(),(2)(1)nfnf nanf n。已知条件中既有nS还有na，有时先求nS，再求na；有时也可直接求na。若1()nnaaf n求na用累加法：11221()()()nnnnnaaaaaaa1a(2)n。已知1()nnaf na求na，用累乘法：121121nnnnnaaaaaaa

21、a(2)n。已知递推关系求na，用构造法（构造等差、等比数列）。特别地，（1）形如1nnakab、1nnnakab（,k b为常数）的递推数列都可以用待定系数法转化为公比为k的等比数列 后，再求na；形如1nnnakak的递推数列都可以除以nk得到一个等差数列后，再求na。（2）形如11nnnaakab的递推数列都可以用倒数法求通项。（3）形如1knnaa的递推数列都可以用对数法求通项。（7）（理科）数学归纳法。（8）当遇到qaadaannnn1111或时，分奇数项偶数项讨论，结果可能是分段形式。数列求和的常用方法：（1）公式法：等差数列求和公式；等比数列求和公式。（2）分组求和法：在直接运用

22、公式法求和有困难时，常将“和式”中“同类项”先合并在一起，再运用公式法求和。（3）倒序相加法：若和式中到首尾距离相等的两项和有其共性或数列的通项与组合数相关联，则常可考虑选用倒序相加法，发挥其共性的作用求和（这也是等差数列前n和公式的推导方法）.（4）错位相减法：如果数列的通项是由一个等差数列的通项与一个等比数列的通项相乘构成，那么常选用错位相减法（这也是等比数列前n和公式的推导方文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4

23、 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2

24、J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4

25、 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2

26、J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4

27、 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2

28、J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4

29、 ZM1C7H9I3R3文德教育5 法）.（5）裂项相消法：如果数列的通项可“分裂成两项差”的形式，且相邻项分裂后相关联，那么常选用裂项相消法求和.常用裂项形式有：111(1)1n nnn；11 11()()n nkk nnk；2211111()1211kkkk，211111111(1)(1)1kkkkkkkkk；1111(1)(2)2(1)(1)(2)n nnn nnn；11(1)!(1)!nnnn；2122(1)2(1)11nnnnnnnnn二、解题方法：求数列通项公式的常用方法：1、公式法2、nnaS 求由（时，时，）naSnaSSnnn121113、求差（商）法如：满足aaaannnn

30、121212251122解：naa1122151411时，naaannn2121212215212211时，12122得：nnaann21annnn141221()()练习数列满足，求aSSaaannnnn111534（注意到代入得：aSSSSnnnnn1114又，是等比数列，SSSnnn144naSSnnnn23411时，4、叠乘法例如：数列中，求aaaannannnn1131解：aaaaaannaannnn213211122311，又，aann133 5、等差型递推公式由，求，用迭加法aaf naaannn110()文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I

31、3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5

32、HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I

33、3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5

34、HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I

35、3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5

36、HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I

37、3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文德教育6 naafaafaaf nnn22321321时，两边相加，得：()()()aafff nn123()()()aafff nn023()()()练习数列，求aaaanannnnn111132（）ann12316、等比型递推公式acad cdccdnn 1010、为常数，可转化为等比数列，设axc axnn 1acacxnn 11令，()cxdxdc11是首项为，为公比的等比数列adcadccn111adcadccnn11

38、11aadccdcnn1111练习数列满足，求aaaaannnn11934（）ann84311 7、倒数法例如：，求aaaaannnn11122由已知得：1221211aaaannnn11121aann111121aan为等差数列，公差为11112121annnann21文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码

39、:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7

40、G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码

41、:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7

42、G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码

43、:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7

44、G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文德教育7 2数列求和问题的方法（1）、应用公式法等差、等比数列可直接利用等差、等比数列

45、的前n 项和公式求和，另外记住以下公式对求和来说是有益的。135(2n-1)=n2【例 8】求数列 1，（3+5），（7+9+10），（13+15+17+19），前 n 项的和。解本题实际是求各奇数的和，在数列的前n 项中，共有 1+2+n=)1(21nn个奇数，最后一个奇数为：1+21n(n+1)-12=n2+n-1 因此所求数列的前n 项的和为（2）、分解转化法对通项进行分解、组合,转化为等差数列或等比数列求和。【例 9】求和 S=1（n2-1）+2 （n2-22）+3（n2-32）+n（n2-n2）解 S=n2（1+2+3+n）-（13+23+33+n3）（3）、倒序相加法适用于给定式子

46、中与首末两项之和具有典型的规律的数列，采取把正着写与倒着写的两个和式相加，然后求和。例 10、求和：12363nnnnnSCCnC例 10、解0120363nnnnnnSCCCnC Sn=3n2n-1（4）、错位相减法如果一个数列是由一个等差数列与一个等比数列对应项相乘构成的，可把和式的两端同乘以上面的等比数列的公比，然后错位相减求和例 11、求数列 1，3x，5x2,(2n-1)xn-1前 n 项的和解设 Sn=1+3+5x2+(2n-1)xn-1(2)x=0时，Sn=1(3)当 x0 且 x1 时，在式两边同乘以x 得 xSn=x+3x2+5x3+(2n-1)xn，文档编码:CM2J7K1

47、J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1

48、C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1

49、J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1

50、C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1J2R5 HC4H2R7G8G4 ZM1C7H9I3R3文档编码:CM2J7K1