九类常见递推数列求通项公式方法

九类常见递推数列求通项公式方法

ID:12012416

大小:30.00 KB

页数:11页

时间:2018-07-15

九类常见递推数列求通项公式方法_第1页
九类常见递推数列求通项公式方法_第2页
九类常见递推数列求通项公式方法_第3页
九类常见递推数列求通项公式方法_第4页
九类常见递推数列求通项公式方法_第5页
资源描述:

《九类常见递推数列求通项公式方法》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库

1、九类常见递推数列求通项公式方法递推数列通项求解方法类型一:,,apaq,,p,1nn,1,,思路1(递推法):apaqppaqqpppaqqq,,,,,,,,,,(),,nnnn,,,123,,,,qqn,12nn,,21………。,,,,,paqpp(1),,,,,pap11,,11pp,,,,q思路2(构造法):设,即得,数列,,apa,,,,,,pq,,1,,,,nn,1p1,,,qqn,1是以为首项、为公比的等比数列,则,即a,,a,,paap,,,,,n1n1,,11pp,,,,,,qqn,1。aap

2、,,,n1,,11pp,,,,例1已知数列满足且,求数列的通项公式。a,1aaa,,23a,,,,nnn,11n解:方法1(递推法):,,aaaa,,,,,,,,,,232(23)3222333……,,nnnn,,,123,,33,,nnn,,,211n,12…。,,,,,23(122,,,,,,,2)1223,,,,2112,,?方法2(构造法):设,即,数列是以aa,,,,,2,,3a,3a,,34,,,,nn,1n1nn,,11n,12为首项、为公比的等比数列,则,即。a,,,,3422a,,23nn1

3、类型二:aafn,,()nn,1思路1(递推法):aafnafnfnafnfnfn,,,,,,,,,,,,,,,,(1)(2)(1)(3)(2)(1)nnnn,,,123n,1…。,,afn(),1i,1思路2(叠加法):,依次类推有:、aafn,,,(1)aafn,,,(2)nn,1nn,,12n,1、…、,将各式叠加并整理得,即aafn,,()aafn,,,(3)aaf,,(1),nn,,2321n1i,1n,1。aafn,,(),n1i,1例2已知,,求。a,1aaan,,1nn,1n解:方法1(递推法

4、):aanannannn,,,,,,,,,,,,,(1)(2)(1)nnnn,,,123nnn(1),………。,,,,a[23,,,,,,,(2)(1)]nnnn,12,1i方法2(叠加法):,依次类推有:、、…、aan,,aan,,,1aan,,,2nn,1nn,,12nn,,23nnnnn(1),aan,,,将各式叠加并整理得,。aa,,2aann,,,,,,,n1121n2,,,i221ii2类型三:afna,,()nn,1思路1(递推法):…afnafnfnafnfnfna,,,,,,,,,,,,,,

5、,,(1)(1)(2)(1)(2)(3)nnnn,,,123…。,,,,fff(1)(2)(3),,,,,fnfna(2)(1)1aann,1思路2(叠乘法):,依次类推有:、,,(1),,(2)fnfnaa,1nn,2aaan,2n2、…、,将各式叠乘并整理得…,,(3),f(1)(1)(2)(3),,,,ffffnaaa11n,3,即…。,,,,fnfn(2)(1)afff,,,,(1)(2)(3),,,,,fnfna(2)(1)n1n,1例3已知,,求。a,1aaa,1nnn,1n,1nnnnnn,,,

6、,,,112123解:方法1(递推法):…aaaa,,,,,,,nnnn,,,123nnnnnn,,,,11112。,nn(1),aaaan,1n,2n,32nn,1n,23方法2(叠乘法):,依次类推有:、、…、、,,,,ana4an,1an,1,1n,2nn,32a21a1nnn,,,123n2,将各式叠乘并整理得…,即,,,,,,,43a3annn,,1111212nnn,,,123…。,,,a,,,,n43(1)nn,nnn,,113类型四:apaqa,,nnn,,11思路(特征根法):为了方便,我们

7、先假定、。递推式对应的特征方程am,an,12n,1p,,2为,当特征方程有两个相等实根时,(、为待定系xpxq,,c,,,dacnd,,n,,2,,数,可利用、求得);当特征方程有两个不等实根时、时,xam,an,x1212nn,,11(、为待定系数,可利用、求得);当特征方程的根aexfx,,fam,an,e12n12为虚根时数列的通项与上同理,此处暂不作讨论。a,,n例4已知、,,求。aa,2a,3aaa,,612nnn,,11n22解:递推式对应的特征方程为即,解得、。x,2x,,3xx,,,6xx,

8、,,6012nn,,11设,而、,即aexfx,,a,2a,312n129,e,,ef,,2,91,5nn,,11,解得,即。a,,,,,2(3),,n1233ef,,55,,f,,5,4n类型五:,,aparq,,pq,,0nn,1,,aan,1nn,1思路(构造法):,设,则aparq,,,,,,,,,,nn,1nn,1qq,,p,,,,,qp,,q,,arar,,1n,从而解得。那么是以为首项

当前文档最多预览五页,下载文档查看全文

此文档下载收益归作者所有

当前文档最多预览五页,下载文档查看全文
温馨提示:
1. 部分包含数学公式或PPT动画的文件,查看预览时可能会显示错乱或异常,文件下载后无此问题,请放心下载。
2. 本文档由用户上传,版权归属用户,天天文库负责整理代发布。如果您对本文档版权有争议请及时联系客服。
3. 下载前请仔细阅读文档内容,确认文档内容符合您的需求后进行下载,若出现内容与标题不符可向本站投诉处理。
4. 下载文档时可能由于网络波动等原因无法下载或下载错误,付费完成后未能成功下载的用户请联系客服处理。