2017春高中数学第2章数列2.4等比数列第2课时等比数列的性质课件新人教A版必修5_图文

新课标导学 数 学 必修5 ·人教A版 第二章 数列 2.4 等比数列 第2课时 等比数列的性质 1 课前自主 2 3 课堂典例 课 时 作 课前自主学习 1915 年, 波兰数学家谢尔宾斯基(W.Sierpinski)创造了一个 美妙的“艺术品”,被人们称为谢尔宾斯基三角形,如图所 示. 如果我们来看一看图中那些白色三角形的个数, 并把它们 按面积大小,从小到大依次排列起来,可以得到一列数: 1,3,9,27,81,……我们知道,这些数构成等比数列,那么等比数列 性质呢? 1.回顾学过的等差(等比)数列知识填空: a n +1 (1)已知{an}为等比数列,则对于任意正整数 n,都有 =_ an an (2)已知{an}为等比数列,则对于任意正整数 n、m 都有 =_ am (3) 若{an}为等差数列,则 a5+a6 成等差数列. ①a1+a2,a3+a4,________ am+k ,a + (m,k∈N*)成等差数列. ②am,________ m 2k a11 2 ③a7+a9=a5+________ =________ a8. 2.已知{an}是等比数列. 1 1 (1)试探讨{3an},{ },{|an|},{- an}是等比数列吗? an 2 (2)a1+a2,a3+a4,a5+a6 成等比数列吗?a1a2,a3a4,a5a6 成 +a2+…+ak,ak+1+ak+2+…+a2k,a2k+1+a2k+2+…+a3k 成等比 (3)计算 a5· a11,a3· a13,a1· a15,你发现了什么? 等比数列主要有以下性质: (1)若{an}是公比为 q 的等比数列,c 为非零常数,则数列{ca 且公比不变仍为 q. 1 1 (2)若{an}是公比为 q 的等比数列, 则数列{ }是公比为 的等比 an q 是公比为|q|的等比数列. (3)若数列{an},{bn}是公比分别为 q,q′的等比数列,则数列 qq′的等比数列. (4)若{an}是等比数列,且 m+n=p+q(m,n,p,q∈N*).则 别地,当 m+n=2p 时,am· an=a2 p; 若{an}是有穷数列, 则与首末两项等距离的两项积相等, 且等 即 a1an=a2an-1=…. (5)若{an}为等比数列,公比为 q,则 an=amqn-m(m,n∈N*). (6)若{an}是等比数列,每隔 k(k∈N*)项取出一项,按原来的顺 列仍是等比数列,且公比为 qk+1. (7)在等比数列{an}中, 连续取相邻 k(k∈N*)项的和(或积)构成 的等比数列. (8){an}是等差数列,c 是正数,则数列{can}是等比数列. (9){an}是等比数列,且 an>0,则{logaan}(a>0,a≠1)是等差 3.试举例探究公比为q的等比数列{an},当q>1,q=1,q< 的增减变化规律. 等比数列的单调性 增 (1)当a1>0,q>1或a1<0,0<q<1时,等比数列{an}为递_______ (3)当q=1时,数列{an}是常数列; (4)当q<0时,数列{an}是摆动数列. 减 (2)当a1>0,0<q<1或a1<0,q>1时,等比数列{an}为递_______ 1 1.(2015· 全国Ⅱ文,9)已知等比数列{an}满足 a1= ,a3a5= 4 导学号 54742417 ( C ) A.2 1 C. 2 B.1 1 D. 8 [解析] 由题意可得 1 =a1q= ,选 C. 2 a3a5=a2 4=4(a4-1)?a4=2,所以 a4 q= a1 3 2 .已知 {an}是等比数列,且 an>0 , a2a4 +2a3a5 +a4a6 =25 导学号 54742418 ( A ) A.5 C.15 B.10 D.20 2 [解析] 由等比数列的性质,得 a4a6=a2 5,a2a4=a3, 2 ∴(a3+a5)2=a2 + 2 a a + a 3 3 5 5, =a2a4+2a3a5+a4a6=25, ∴a3+a5=± 5. ∵an>0,∴a3+a5=5. 3.等比数列{an}中,首项为 a1,公比为 q,则下列条件中, 减数列的条件是 导学号 54742419 ( C ) A.|q|<1 B.a1>0,q<1 C.a1>0,0<q<1 或 a1<0,q>1 D.q>1 [解析] 等比数列的增减性由首项的符号以及公比的绝对值 -an=a1qn-1(q-1)<0,得 a1>0,0<q<1,或 a1<0,q>1. 4 .(2016· 全国卷Ⅰ理,15)设等比数列{an}满足 a1 +a3 =10 64 a1a2…an 的最大值为________. 导学号 54742420 [解析] 设{an}的公比为 q,由 a1+a3=10,a2+a4=5 得 a1 1 =4,a3=2,a4=1,a5= ,所以 a1a2…an≤a1a2a3a4=64. 2 课堂典例讲练 命题方向1 ?等比数列的性质 在等比数列{an}中, 已知 a4a7=-512, a3+a8=124 512 则 a10=________. 导学号 54742421 [解析] 由等比数列的性质,得 a3a8=a4a7=-512, ? ?a3+a8=124 由? ? ?a3a8=-512 ,得 ? ?a3=-4 ? ? ?a8=128 ? ?a3=128 或? ? ?a8=-4 . ∵q 为整数,∴a3=-4,a8=128. a8 128 ∴q = = =-32,∴q=-2. a3 -4 5 ∴a10=a8· q2=128×4=512. 1 『规律总结』 (1)若{an}为等比数列, 则{ }, {|an|}, {a2 { n}, an +k }均为等比数列; an (2)若{an},{bn}均为等比数列,则{anbn},{

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