2021年特征方程求递推数列通项公式.pdf

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1、特征方程求递推数列通项公式一、一阶线性递推数列通项公式的研究与探索若数列na满足),1(,11cdcaabann求数列na的通项na它的通项公式的求法一般采用如下的参数法，将递推数列转化为等比数列：设tccaatactannnn)1(),(11则，令dtc)1(，即cdt1，当1c时可得)1(11cdaccdann，知数列cdan1是以c为公比的等比数列，将ba1代入并整理，得11cdcbdbcannn.观察可发现cd1即为方程dcxx的根我们称方程dcxx为递推公式)1(1cdcaann的特征方程，cd1为特征方程的根。将上述参数法类比到二阶线性递推数列,11nnnqapaa能得到什么结论？

2、二、二阶线性递推数列通项公式的研究与探索若数列na满足,11nnnqapaa设)(11nnnntaastaa，则11)(nnnstaatsa,令qstpts（1）若方程组有两组不同的实数解),(),(2211tsts,则)(11111nnnnatasata,)(12221nnnnatasata,即nnata11、nnata21分别是公比为1s、2s的等比数列，精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 1 页，共 4 页由等比数列性质可得1111211)(nnnsataata,1212221)(1nnnsataata,21tt由上两式消去1na可得nnnsttsatast

3、tsataa22121221211112.（2）若方程组有两组相等的解2121ttss，易证此时11st，则)(2112111111nnnnnnatasatasata)(11211atasn，211121111sasasasannnn,即nnsa1是等差数列，由等差数列性质可知21112111.1sasansasann，所以nnsnsasasasasaa1211122111211.这样，我们通过参数方法，将递推数列转化为等比（差）数列，从而求得二阶线性递推数列的通项，若将方程组消去t即得02qpss，显然1s、2s就是方程qpxx2的两根，我们不妨称此方程为二阶线性递推数列11nnnqapaa

4、的特征方程，结论：设递推公式为,11nnnqapaa其特征方程为022qpxxqpxx即，1、若方程有两相异根1s、2s，则nnnscsca2211；2、若方程有两等根21ss，则nnsncca121)(.其中1c、2c可由初始条件确定。例 1.（1）已知数列na满足*12212,3,32()nnnaaaaanN，求数列na的通项na（2）已知数列na满足*12211,2,44()nnnaaaaanN，求数列na的通项na精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 2 页，共 4 页文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:

5、CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P

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9、CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P

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11、CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8三、分式线性递推数列dacbaaannn 1通项公式的研究与探索仿照前面方法，等式两边同加参数t，则dacctadtbactatdacbaatannnnn)(1令ctadtbt，即0)(2btdact记此方程的两根为21,tt，（1）若21tt，将21,tt分别代入式可得以上两式相除得21212111tatactactatatannnn，于是得到21tatann为等比数列，其公比为21ctacta，数列na的通项na可由121211121)(nn

12、nctactatatatata求得；（2）若21tt，将1tt代入式可得dactactatannn1111)(,考虑到上式结构特点，两边取倒数得111111)(11tactdtacctatannn由于21tt时方程的两根满足cdat12，11ctdcta于是式可变形为111111tactactann11tan为等差数列，其公差为1ctac，数列na的通项na可由1111)1(11ctacntatan求得精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 3 页，共 4 页文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V

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16、8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V

17、7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X

18、8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V

19、7 HD4W8C4P3R3 ZF2R3X8I3D8文档编码:CP5K7O2C4V7 HD4W8C4P3R3 ZF2R3X8I3D8这样，利用上述方法，我们可以把分式线性递推数列转化为等比数列或等差数列，从而求得其通项。如果我们引入分式线性递推数列dacbaaannn 1（0,cRdcba）的特征方程为dcxbaxx，即0)(2bxadcx，此特征方程的两根恰好是方程两根的相反数，于是我们又有如下结论：分式线性递推数列dacbaaannn 1（0,cRdcba），其特征方程为dcxbaxx，即0)(2bxadcx，1、若方程有两相异根1s、2s，则21sasann成等比数列，其公比为21csac

20、sa；2、若方程有两等根21ss，则11san成等差数列，其公差为1csac.例 2.（1）已知数列na满足*1112,2,nnaanNa,求通项na.（2）已知数列na满足11122,(2)21nnnaaana，求数列na的通项na（3）设数列na满足nnnnaaaaa求,7245,211.例 3.（09江西 理22）各项均为正数的数列na,12,aa ab,且对满足mnpq的正数,m n p q都有(1)(1)(1)(1)pqmnmnpqaaaaaaaa.(1)当14,25ab时,求通项na;精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 4 页，共 4 页文档编码:CP

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