等比数列 复习资料题
数列专题(三):等比数列
知识点 等比数列的基本概念和等差数列的区别与联系
1. 等比数列 等差数列
aa 定义: n?1?q或n?q an?1?an?d或an?an?1?danan?1
公比: q 公差: d
?递增数列 :a1?0, q?1?递增数列:d?0
单调性: ?a?0,则反之 ?1?? 递减数列 :a?0, 0?q?1?递减数列:d?0?1 通项公式: an?a1qn?1 an?a1??n?1?d
?①等差中项:若a,A,b呈是等差数列?①等比中项:若a,A,b成等比数列 ??a?b?2 性质: ? 则 A?ab ? 则 A?2?②若m?n?p?q,则a?a?a?a? mnpq???②若m?n?p?q,则am?an?ap?aq
??①定义法:an?an?1?dn?2,且n?N*或an?1?an?d?数列?an?为等差数列?? ??*?1?等差数列的判定:?②等差中项法:2an?an?1?an?1(n?2,且n?N)?数列?an?为等差数列 ??③通项公式法:an?kn?b(k,b为常数)?数列?an?为等差数列?? ??2. ?anan?1?*①定义法:?qn?2,且n?N或?q ?数列?an?为等比数列?? ??2?等比数列的判定:aan?1n? ??②等比中项法:a2?a?a(n?2,且n?N*)?数列?a?为等比数列nn?1n?1n??
**注意:①a?a?d(d为常数,n?N)对任意的n?N恒成立,不能几项成立就说?an?为等差数列。n?1n
②an?1?q(q为常数,n?N*)对任意的n?N*恒成立,不能几项成立就说a为等比数列。 ?n?a n ??①若是两个数呈等差数列,则可设为a?d,a?d;?? 1等差数列的假设???②若是三个数呈等差数列,则可设为a?d,a,a?d;? ?③若是四个数呈等差数列,则可设为a?3d,a?d,a?d,a?3d.???
??a ?①若是两个数呈等比数列,则可设为,aq;?3.类比?q ????2?等比数列的假设?②若是三个数呈等比数列,则可设为a,a,aq;
??q ???aa ?③若是四个数呈等比数列,则可设为,,aq,aq3.?3?qq ??????
考点一 等比数列的通项公式:利用方程的思想求出等比数列的首项a1和公比q ?an?a1qn?1例1 ?1(?2013?北京高考)等比数列?an?满足a2?a4?20,a3?a5?40,则公比q?_________24??a1?2?a1q?a1q?20 ①方程①?q??a1q?a1q?20q 解:???????2?2?44aq?aq?40 ②aq?aq?40???q?2?11?11
3
等比数列 复习资料题
? 2(?2014?江苏高考)已知等比数列?an?的各项均为正数,且a2?1,a8?a6?2a4,则 a6?________解析:①运用解方程的思想,求首项a1和公比q②若求出首项a1和公比q很麻烦,数字很大或很难处理时,有时需要整体代换2?q?2?解:a8?a6?2a4?a1q7?a1q5?2a1q3?q4?q2?2?q4?q2?2?0??2?a6?a2q4?4??q??1?舍?
强化练习:1 已知等比数列?an?的公比为正数,且a2?a6?9a4,a2?1,则 a1?? ?11 A. 3 B. ?3 C. ? D. 332 已知等比数列?an?中,且a6?a2?34,a6?a2?30,则 a4?? ? A. 8 B. 16 C. ?8 D. ?1623 已知等比数列?an?中,满足a1?2,a3a5?4a6,则 a3?? ?11 A. B. 1 C. 2 D. 244 已知等比数列?an?中,且a1?a2?324,a3?a4?36,则a5?a6?________5 已知等比数列?an?中,且a5?a6?27,a7?a8?81,则a3?a4?________
? ?①等比中项:a,G,b成等比数列,则G?ab;考点二 等比数列的性质? ??②若m?n?p?q,则aman?ap?aq
例2 ?2014?天津高考?设?an?是首项为a1,公差为?1 的等差数列,Sn为其前n项和,若
S1,S2,S4成等比数列,则a1?? ?
11 A 2 B ?2 C D ?22
解析:利用等比中项的性质。
S,S,S成等比数列?S2?S?S ??2a?d?2?a?4a?6d?,代入d??1解得a??1?12421411112
例 3 ?2014?广东高考?等比数列?an?的各项均为正数,且a1a5=4, log2a1?log2a2?log2a3?log2a4?log2a5?____________则
?lne?1A?
解析:考察知识点①lgA?lgB?lgAB; ②lgA?lgB?lg; ③lgAB?BlgA; ④?lg10?1, B ?log1?0?a ②若m?n?p?q,则aman?ap?aq log 2a1?log2a2?log2a3?log2a4?log2a5?log2a1a2a3a4a5,又a1a5?a2a4?a3a3?4? log2a1?log2a2?log2a3?log2a4?log2a5?log2a1a2a3a4a5?log24?2?4?log232?52等比数列 复习资料题
强化练习: 1(?? 2014?全国高考II)等差数列?an?公差为2,若a2,a4,a8成等比数列,则?an?的前n项和Sn?? ?n?n?1?n?n?1? A. n(n?1) B. n?n?1? C. D. 22
6x?6,...的第四项等于? ??2(? 2013?江西高考)等比数列x,3x?3, A. ?24 B. 0 C. 12 D. 24
?3(? 2014?安徽高考)数列?an?是等差数列,若a1?1,a3?3,a5?5构成公比为q的等比数列,则q?___ ?4(?2014?山东高考)等差数列?an?中,已知公差d?2,若a2是a1与a4的等比中项,则an?___________ ?5(? 2014?山东高考)等差数列?an?的公差d?2,前n项和为Sn,且S1,S2,S4成等比数列,则an?_____ ?6(? 2014?重庆高考)对任意等比数列?an?,下列说法一定正确的是? ? A. a1,a3,a9成等比数列 B. a2,a3,a6成等比数列 C. a2,a4,a8成等比数列 D. a3,a6,a9成等比数列
?7?等比数列?an?的各项均为正数,且a1a9=2,则log2a3?log2a4?log2a5?log2a6?log2a7?_______
?8?等比数列?an?的各项均为正数,且a2a7=10,则lga1?lga3?lga4?lga5?lga6?lga8?__________
?9?等比数列?an?的各项均为正数,且a2a7=10,则lga1?lga3?lga4?lga5?lga6?lga8?__________ ?10(? 2014?广东高考)等比数列?an?的各项均为正数,且a10a11?a9a12?2e5,则
lna1?lna2?...?lna20?_____________ ?11(? 2014?全国高考)等比数列?an?中,已知a4?2,a5?5,则数列?lgan?的前8项和等于 ? ? A. 6 B. 5 C. 4 D. 3
三 等比数列的判定考点 anan?1?*①定义法:?qn?2,且n?N或?q?数列?an?为等比数列? aa等比数列的判定:n?1n? ?②等比中项法:a2?a?a(n?2,且n?N*)?数列?a?为等比数列nn?1n?1n?
注意:在说明一个数列是等比数列的同时,必须交代首项和公比分别是什么。
*例4 ?1?已知数列?an?中,an?2an?1n?2,且n?N,且a1?1,则通项公式an?__________ ?2?已知各项为正数的数列?an?中,an?1?1?3?an?1?,且a1?3,则通项公式an?__________ 解析:?1?可用定义法直接判定数列?an?为等比数列;
?2?以新数列的视界看待?an?1?,数列?an?1?是以a1?1?2为首项,公比为3的等比数列。
a解:2为公比的等比数列,即a?aqn?1?2n?1? 1??a?2a?n?2 ?数列?a?是以a?1为首项,????nn?1an?1n1n1等比数列 复习资料题
强化练习: an?1?1?3.即?an?1?是以a1?1?2为首项,公比为3的等比数列 an?1 ?a?1?2?3n?1?a?2?3n?1?1 nn
例5 已知数列?an?的前n项和为Sn,且an?Sn?n
?1 ?设cn?an?1,求证:?cn?是等比数列 ?2?求数列?an?的通项公式.c 解析:思路由Sn?Sn?1?an?cn?n?q??cn?是等比数列??an?的通项公式cn?1
?1?证明:解:?S1?a1 ①?②得an?an?1?an?1 1 ?a1?a1?1?a1? ?2an?an?1?1?2?an?1??an?1?1 2 a?11 ?an?Sn?n................① ?n? an?1?12 11 ?a?S?n?1........② ?c是以a?1??为首项,公比为的等比数列 ??n?1n?1n1 22n?1nn 1?1??1??1? ?2??cn??2??2????2? ,又cn?an?1? cn?1?an ?an???2??1??????
?2??an?1?1?3?an?1???1?已知数列?an?中,an?1an?1?n?2,且n?N*?,且a1?2,则通项公式an?__________22,则通项公式an?________________3?2?已知数列?an?中,an?1?3an?n?N*?,且a1??3?已知各项为正数的数列?an?中,an?1?1?2?an?1?,且a1?3,则通项公式an?__________?4?已知各项为正数的数列?an?中,an?1?12?1???an??,且a1?1,则通项公式an?__________23?2?21an?,则数列?an?的通项公式an?________________33?6?已知数列?an?的前n项和为Sn,且Sn?4an?3?n?N??,?5??2013?全国卷I?若数列?an?的前n项和Sn? ?I?证明:数列?an?是等比数列. ?II?求?an?的通项公式.?7??2014?全国高考?已知数列?an?满足a1?1,an?1?3an?1?n?N??,1?? ?I?证明:数列?an??是等比数列. ?II?求?an?的通项公式. 2??
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