书签分享收藏举报版权申诉 / 6

立即下载

当前位置：首页 > 教育专区 > 高考资料 > (完整word版)高考文科数学数列复习题.pdf

(完整word版)高考文科数学数列复习题.pdf

上传人：H****o

文档编号：55056184

上传时间：2022-10-29

格式：PDF

页数：6

大小：336KB

( 4.5 )

《(完整word版)高考文科数学数列复习题.pdf》由会员分享，可在线阅读，更多相关《(完整word版)高考文科数学数列复习题.pdf（6页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、-来源网络，仅供参考高考文科数学数列复习题一、选择题1已知等差数列共有10 项，其中奇数项之和15，偶数项之和为30，则其公差是（）A5 B4 C3 D2 2在等差数列na中，已知1232,13,aaa则456aaa等于（）A40 B 42 C43 D 45 3已知等差数列na的公差为2，若1a、3a、4a成等比数列，则2a等于（）A 4 B 6 C 8 D 10 4.在等差数列na中,已知11253,4,33,naaaan则为()A.48 B.49 C.50 D.51 5在等比数列na 中，2a 8，6a64，则公比q为（）A2 B3 C4 D8 6.-1,a,b,c,-9成等比数列，那么

2、（）A3,9bac B.3,9bac C.3,9bac D.3,9bac7数列na满足11,(2),nnna aan na则（）A(1)2n n B.(1)2n n C.(2)(1)2nn D.(1)(1)2nn8已知abcd，成等比数列，且曲线223yxx的顶点是()bc，则ad等于（3 2 1 29在等比数列na中，12a，前n项和为nS，若数列1na也是等比数列，则nS等于（）A122nB3nC2nD31n10设4710310()22222()nf nnN，则()f n等于（）A2(81)7nB12(81)7n C 32(81)7nD42(81)7n二、填空题（5 分 4=20 分）11

3、.已知数列的通项52nan，则其前n项和nS12已知数列na对于任意*pqN，有pqpqaaa，若119a，则36a13数列an中，若a1=1，2an+1=2an+3（n1），则该数列的通项an=.14已知数列na是首项为1，公差为 2 的等差数列，将数列na中的各项排成如图所示的一个三角形数表，记A（i,j)表示第 i 行从左至右的第j 个数，例如A（4，3）=9a,则 A（10，2）=三、解答题（本大题共6 题，共 80 分，解答应写出文字说明、证明过程或演算步骤）15、（本小题满分12 分）等差数列的通项为219nan,前 n 项和记为ns，求下列问题:(1)求前 n 的和ns（2）当

4、n 是什么值时,ns有最小值，最小值是多少？16、（本小题满分12 分）-来源网络，仅供参考数列na的前n项和记为nS，111,211nnaaSn（1）求na的通项公式;（2）求nS17、（本小题满分14 分）已知实数列是na等比数列,其中74561,1,aaaa且成等差数列.(1)求数列na的通项公式;(2)数列na的前n项和记为,nS证明:nS128,3,2,1(n).18、（本小题满分14 分）数列na中，12a，1nnaacn（c是常数，12 3n，），且123aaa，成公比不为1的等比数列（1）求c的值；（2）求na的通项公式19、（本小题满分14 分）设na是等差数列，nb是各项都

5、为正数的等比数列，且111ab，3521ab，5313ab（1）求na，nb的通项公式；（2）求数列nnab的前n项和nS20（本小题满分14 分）设数列na满足211233333nnnaaaa，a*N（1）求数列na的通项；（2）设nnnba，求数列nb的前n项和nS1.（本题满分14 分）设数列na的前n项和为nS,且34nnaS(1,2,)n，（1）证明:数列na是等比数列；（2）若数列nb满足1(1,2,)nnnbabn，12b，求数列nb的通项公式2.（本小题满分12 分）等比数列na的各项均为正数，且212326231,9.aaaa a1.求数列na的通项公式.2.设31323lo

6、glog.log,nnbaaa求数列1nb的前项和.3.设数列na满足21112,3 2nnnaaa（1）求数列na的通项公式；（2）令nnbna，求数列的前n 项和nS文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:

7、CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C

8、10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4

9、 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4

10、F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6

11、 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D1

12、0E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C

13、8-来源网络，仅供参考4.已知等差数列an 的前 3 项和为 6，前 8 项和为 4（）求数列an 的通项公式；（）设bn=（4an）qn1（q0，nN*），求数列 bn 的前 n 项和 Sn5.已知数列 an 满足，nN（1）令 bn=an+1an，证明：bn 是等比数列；（2）求 an 的通项公式高三文科数学数列测试题答案15 CBBCA 610 BABCD 11.(51)2nn 12.4 13.3122nan 14.93 15.略解（1）略（2）由100nnaa得10n，10 910210(17)2260s16.解：（1）设等比数列na的公比为()q qR，由6711aa q，得61aq

14、，从而3341aa qq，4251aa qq，5161aa qq因为4561aaa，成等差数列，所以4652(1)aaa，即3122(1)qqq，122(1)2(1)qqq所以12q故116111642nnnnaa qqq（2）1164 12(1)1128 112811212nnnnaqSq17（1）由121nnaS可得1212nnaSn，两式相减得112,32nnnnnaaaaan又21213aS213aa故an是首项为1，公比为3 得等比数列13nna.(2)1(1 3)311 322nnnS18.解：（1）12a，22ac，323ac，因为1a，2a，3a成等比数列，所以2(2)2(23

15、)cc，解得0c或2c当0c时，123aaa，不符合题意舍去，故2c（2）当2n时，由于21aac，322aac，1(1)nnaanc，所以1(1)12(1)2nn naancc又12a，2c，故22(1)2(2 3)nan nnnn，当1n时，上式也成立，所以22(1 2)nannn，文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6

16、ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10

17、E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8

18、文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:C

19、K5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C1

20、0N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4

21、HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F

22、4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8-来源网络，仅供参考19.解：（1）设na的公差为d，nb的公比为q，则依题意有0q且4212211413dqdq，解得2d，2q所以1(1)21nandn，112nnnbq（2）1212nnnanb122135232112222nnnnnS，3252321223222nnnnnS，得22122221222222nnnnS，1111212221212nnn12362nn20(1)2112333.3,3nnnaaaa1.解：（1）证：因为34nnaS(1,2,)n，则3411n

23、naS(2,3,)n，所以当2n时，1144nnnnnaSSaa，整理得143nnaa 5分由34nnaS，令1n，得3411aa，解得11a所以na是首项为1，公比为43的等比数列 7分（2）解：因为14()3nna，由1(1,2,)nnnbabn，得114()3nnnbb 9分由累加得)()()(1231 21nnnbbbbbbbb1)34(3341)34(1211nn，（2n），当 n=1 时也满足，所以1)34(31nnb2.解：（）设数列an 的公比为q，由23269aa a得32349aa所以219q。有条件可知a0,故13q。由12231aa得12231aa q，所以113a。故

24、数列 an 的通项式为an=13n。文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6

25、ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10

26、E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8

27、文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:C

28、K5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C1

29、0N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4

30、HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8-来源网络，仅供参考（?）111111loglog.lognbaaa故12112()(1)1nbn nnn所以数列1nb的前 n 项

31、和为21nn3.解：（）由已知，当n1 时，2(1)12n。而12,a所以数列 na的通项公式为212nna。（）由212nnnbnan知35211 22 23 22nnSn从而23572121 22 23 22nnSn-得2352121(12)22222nnnSn。即211(31)229nnSn4.解：（1）设 an的公差为d，由已知得解得 a1=3，d=1 故 an=3+（n1）（1）=4n；（2）由（1）的解答得，bn=n?qn 1，于是Sn=1?q0+2?q1+3?q2+（n1）?qn1+n?qn若 q1，将上式两边同乘以q，得qSn=1?q1+2?q2+3?q3+（n1）?qn+n?

32、qn+1将上面两式相减得到（q1）Sn=nqn（1+q+q2+qn 1）=nqn文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5

33、Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N

34、9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB

35、8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z

36、5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM

37、7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9

38、M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8-来源网络，仅供参考于是 Sn=若 q=1，则 Sn=1+2+3+n=所以，Sn=5.解

39、：（1）证 b1=a2a1=1，当 n2 时，所以 bn 是以 1 为首项，为公比的等比数列（2）解由（1）知，当 n2 时，an=a1+（a2a1）+（a3a2）+（anan 1）=1+1+（）+=，当 n=1 时，所以文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8

40、Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5

41、E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7

42、D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M

43、1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编

44、码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z

45、8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8文档编码:CK5Z8C10N9A4 HB8Y4F4Z5E6 ZM7D10E9M1C8

文档加载中……请稍候！
如果长时间未打开，您也可以点击刷新试试。

下载文档到电脑，查找使用更方便

4.3 金币

版权申诉 word格式文档无特别注明外均可编辑修改；预览文档经过压缩，下载后原文更清晰！ 立即下载

配套讲稿：: 如PPT文件的首页显示word图标，表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
特殊限制：: 部分文档作品中含有的国旗、国徽等图片，仅作为作品整体效果示例展示，禁止商用。设计者仅对作品中独创性部分享有著作权。
关键词：: 完整 word 高考文科数学数列复习题

淘文阁 - 分享文档赚钱的网站所有资源均是用户自行上传分享，仅供网友学习交流，未经上传用户书面授权，请勿作他用。

限制150内

关于本文

本文标题：(完整word版)高考文科数学数列复习题.pdf
链接地址：https://www.taowenge.com/p-55056184.html