(本题满分18分,第1小题4分,第2小题6分,第3小题8分)已知数列{ a n }满足 , (其中λ≠0且λ≠–1
(本题满分18分,第1小题4分,第2小题6分,第3小题8分)已知数列{an}满足,(其中λ≠0且λ≠–1,n∈N*),为数列{an}的前项和.(1)若,求的值;(2)求数...
(本题满分18分,第1小题4分,第2小题6分,第3小题8分)已知数列{ a n }满足 , (其中λ≠0且λ≠–1, n ∈N*), 为数列{ a n }的前 项和.(1) 若 ,求 的值;(2) 求数列{ a n }的通项公式 ;(3) 当 时,数列{ a n }中是否存在三项构成等差数列,若存在,请求出此三项;若不存在,请说明理由.
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灰哥哥囤妛僺溟
2014-10-11
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(1) ![](https://iknow-pic.cdn.bcebos.com/d6ca7bcb0a46f21f5a7936e0f5246b600c33ae30?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ;(2)数列{ a n }中存在 a 1 、 a 2 、 a 3 或 a 3 、 a 2 、 a 1 成等差数列。 |
试题分析:(1) 令 ![](https://iknow-pic.cdn.bcebos.com/6c224f4a20a44623d287cdd19b22720e0cf3d730?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,得到 ![](https://iknow-pic.cdn.bcebos.com/8694a4c27d1ed21b7d6b4656ae6eddc450da3fe6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,令 ![](https://iknow-pic.cdn.bcebos.com/d000baa1cd11728b218fe491cbfcc3cec2fd2ce6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,得到 ![](https://iknow-pic.cdn.bcebos.com/cc11728b4710b912a0639e4dc0fdfc03934522e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 。…………2分 由 ![](https://iknow-pic.cdn.bcebos.com/730e0cf3d7ca7bcbe836af9cbd096b63f624a830?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,计算得 ![](https://iknow-pic.cdn.bcebos.com/d6ca7bcb0a46f21f5a7936e0f5246b600c33ae30?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) .……………………………………………………4分 (2) 由题意 ![](https://iknow-pic.cdn.bcebos.com/4610b912c8fcc3cea962a1809145d688d53f20e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,可得: ![](https://iknow-pic.cdn.bcebos.com/c9fcc3cec3fdfc03feda8b0bd73f8794a5c226e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,所以有 ![](https://iknow-pic.cdn.bcebos.com/c2fdfc039245d688baa0da17a7c27d1ed31b24e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,又 ![](https://iknow-pic.cdn.bcebos.com/fd039245d688d43fe80bf9417e1ed21b0ff43be6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,……………………5分 得到: ![](https://iknow-pic.cdn.bcebos.com/eaf81a4c510fd9f9395a3dda262dd42a2834a430?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,故数列 ![](https://iknow-pic.cdn.bcebos.com/d53f8794a4c27d1ea084537718d5ad6edcc438e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 从第二项起是等比数列。……………7分 又因为 ![](https://iknow-pic.cdn.bcebos.com/8694a4c27d1ed21b7d6b4656ae6eddc450da3fe6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,所以 n ≥2时, ![](https://iknow-pic.cdn.bcebos.com/d31b0ef41bd5ad6eab5b0c5982cb39dbb7fd3ce6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ……………………………8分 所以数列{ a n }的通项 ![](https://iknow-pic.cdn.bcebos.com/0ff41bd5ad6eddc42645dc483adbb6fd536633e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) …………………………………10分 (3) 因为 ![](https://iknow-pic.cdn.bcebos.com/ac6eddc451da81cb4044eb7e5166d016082431e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 所以 ![](https://iknow-pic.cdn.bcebos.com/dcc451da81cb39dbcc620fe5d3160924aa1830e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ……………………………………11分 假设数列{ a n }中存在三项 a m 、a k 、a p 成等差数列, ①不防设 m > k > p ≥2,因为当 n ≥2时,数列{ a n }单调递增,所以2 a k =a m +a p 即:2´( ![](https://iknow-pic.cdn.bcebos.com/023b5bb5c9ea15cecea388bab5003af33a87b230?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) )´4 k –2 = ![](https://iknow-pic.cdn.bcebos.com/023b5bb5c9ea15cecea388bab5003af33a87b230?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ´4 m –2 + ![](https://iknow-pic.cdn.bcebos.com/023b5bb5c9ea15cecea388bab5003af33a87b230?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ´4 p –2 ,化简得:2´4 k - p = 4 m – p +1 即2 2 k– 2 p +1 =2 2 m –2 p +1,若此式成立,必有:2 m – 2 p =0且2 k –2 p +1=1, 故有: m=p=k ,和题设矛盾………………………………………………………………14分 ②假设存在成等差数列的三项中包含 a 1 时, 不妨设 m =1, k > p ≥2且 a k > a p ,所以2 a p = a 1 + a k , 2´( ![](https://iknow-pic.cdn.bcebos.com/023b5bb5c9ea15cecea388bab5003af33a87b230?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) )´4 p –2 = – + ( ![](https://iknow-pic.cdn.bcebos.com/023b5bb5c9ea15cecea388bab5003af33a87b230?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) )´4 k –2 ,所以2´4 p– 2 = –2+4 k –2 ,即2 2 p –4 = 2 2 k –5 – 1 因为 k > p ≥ 2,所以当且仅当 k =3且 p =2时成立………………………………………16分 因此,数列{ a n }中存在 a 1 、 a 2 、 a 3 或 a 3 、 a 2 、 a 1 成等差数列……………………………18分 点评:本题主要考查了利用数列的递推公式求解数列的通项公式,还考查了一定的逻辑运算与推理的能力及考查了学生通过已知条件分析问题和解决问题的能力.题目较难。 |
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