高三数学-理科数列-专题练习(含答案与解析)

高三数学专题练习 理科数列 一、选择填空. 1.已知对任意n?N+都有an?n(n??)恒成立,且数列{an}是递增数列,则实数?的取值范围是( ) A.(?,+?) 72 B.(?3,??) C.(??,?3) D.[?2,+?) 2.已知数列{an}和数列{bn}满足a1?,b1?A.{an}是递增数列,{bn}是递减数列 B.{an}是递减数列,{bn}是递增数列 C.{a2n}是递增数列,{bn}是递减数列 D.{a2n?1}是递减数列,{bn}是递减数列 431,且an?1?2bn?3,bn?1?2an?4,则( ) 33.设数列{an}的前n项积为Tn,则满足不等式k?T12?T22?A.1 B.2 C.3 12?Tn2的最小整数k为( ) D.4 4.已知Sn是数列{an}的前n项和,a1?1,a2?2,a3?3,数列{an?an?1?an?2}是公差为2的等差数列,则S25?( ) A.232 B.233 C.234 D.235 sin2a3?sin2a7??1,当n?10时,数列{an}的前n项和Sn取得最5.已知等差数列{an}的公差d?(0,1),且sin(a3?a7)小值,则首项a1的取值范围是( ) A.(?5π9π,?) 816 B.[?5π9π,?] 816 C.(?5π9π,?) 48 D.[?5π9π,?] 486.已知数列{an}满足an?2?an?1?an,n?N*,a1?a,a2?b,其前n项和为Sn,则S2016? . 7.设Sn为数列{an}的前n项和,Sn?二、解答题. 1.已知{an}是各项均为正数的等比数列,{bn}是等差数列,a1?b1?1,b2?b3?2a3,a5?3b2?7. (Ⅰ)求{an}和{bn}的通项公式； 1n*?(?1)a(n?N),则数列{Sn}的前9项和为 . nn2 - 1 - / 8

(Ⅱ)设cn?bn,n?N*,求证：c1?c2?an?cn?6. 2.数列{an}的前n项和为Sn,a1?1,an?1??,Sn?1(???1,n?N*)且a1,2a2,a3?3成等差数列. (Ⅰ)求an； (Ⅱ)设bn?log4an?1,Cn?1bn?bn?1?bn?2,求证：c1?c2??cn?2. 3.设正项数列{an}的前n项和为Sn,an,Sn,21成等差数列. an(Ⅰ)证明{Sn}是等差数列,并求{an}的通项公式； (Ⅱ)证明2(n?1?1)?11??S1S2?1?2n?1. Sn4.在单调递增数列{an}中,a1?1,a2?2,且a2n?1,a2n,a2n?1成等差数列,a2n,a2n?1,a2n?2成等比数列. (Ⅰ)求数列{an}的通项公式； (Ⅱ)设数列{1}的前n项和为Sn,证明：Sn?4n. ann?225.数列{an}满足an?1?an?nan?1,当a1?3时,求证： (Ⅰ)an?n?2； (Ⅱ)11??1?a11?a2?11?. 1?an2 - 2 - / 8

高三数学专题练习

理科数列 答案

一、选择填空 1.B 22【解析】∵{an}是递增数列,∴an?1?an,即(n?1)??(n?1)?n??n∴???2n?1对于n?N+恒成立,∴???3,故选B. 2.D 【解析】∵an?1?2bn?3,bn?1?2an?4, ∴an?1?bn?1?2(a?b)?1, 又a1?,b1?431?an?bn?1?an?bn?1 3∴{an}与{bn}的单调性一样,故选D. 3.B 【解析】∵Tn?1?1?an?1,an?1?Tn?11?an?111???1, ,整理得Tn1?an1?an?11?an111?2?n?1,∴an?n,∴Tn?1, 且a1?,∴,∴1?a11?ann?12n?1T12?T22??Tn2?11111??????2232(n?1)21223111???1??1nn?1n?1?1n(n?1)111?1????22312 ∴k?1,∴k?2. 4.B 【解析】由题前25项的和可以看作第1项加上以第2,3,4项为首项,三个公差为2的等差数列的前8项之和. 由题可得a4?3,∴a2?a3?a4?8,∴S25?1?2?8?故选B. 5.D 8?78?78?7?2?3?8??2?3?8??2?233, 222 3 / 8

sin2a3?sin2a7cos2a7?cos2a3??1转化为??sin(a3?a7)再利用【解析】利用三角函数的降幂公式将条件sin(a3?a7)2和差化积公式转化,求得sin(a7?a3)?1,从而可求得等差数列{an}的公差d??a10?0π,根据?即可求得首项a?08?11a1的取值范围. sin2a3?sin2a7??1, ∵{an}为等差数列,sin(a3?a7)1?cos2a31?cos2a7?cos2a7?cos2a3??sin(a3?a7), ∴,∴22??12sin(a3?a7)∴?(?2)sin(a7?a3)?sin(a7?a3)??sin(a3?a7),∴sin(a3?a7)?0,∴sin(a7?a3)?1, ∴4d?2kπ?12πππ?(0,4),∴k?0,∴4d?,∴d?, 282π?a?9??0??a10?0?15π9π8?a1??, ∵n?10时,数列{an}的前n项和Sn取得最小值,?,∴?,∴?48?a11?0?a?10?π?01?8?故选D. 6.0 【解析】由递推公式an?2?an?1?an,n?N*可得数列的前几项为:a,b,b?a,?a,?b,a?b,a,b,, 由此可得数列是周期为6的周期数列. 又因为a?b?(b?a)?(?a)?(?b)?(a?b)?0,且2016?6?336 所以S20167.??0. 341 1024【解析】 11?a,S??, 2k2k?122k22k1*当n?2k?1(k?N)时,S2k?1?2k?1??a2k?1 2111∴a2k?1?2k?2k?1??2k,∴S2k?2?S2k?1?a2k?1?0(k?2), 222当n?2k(k?N)时,S2k?* 4 / 8