2.4.2 等比数列的性质（学案）-【成才之路】2020-2021学年高中新课程数学同步学习指导（人教A版必修5）

数学 （必修 5·人教 A 版）　 因为 b≠c,所以 c = 4b,所以 a = - 2b, 代入 a + 3b + c = 10 中解得 b = 2,所以 a = - 4. 　 　 典例试做 3:(1)证明:∵ an +1 =2an +1,∴ an +1 +1 =2(an +1), 即 bn +1 =2bn, ∵ b1 = a1 + 1 = 2≠0. ∴ bn ≠0,∴ bn + 1 bn = 2,∴ {bn }是等比数列. (2)由(1)知{bn }是首项 b1 = 2,公比为 2 的等比数列, ∴ bn = 2 × 2 n - 1 = 2n ,即 an + 1 = 2 n ,∴ an = 2 n - 1. 　 　 跟踪练习 3:(1)由 S1 = 1 3 (a1 - 1), 得 a1 = 1 3 (a1 - 1), 所以 a1 = - 1 2 , 又 S2 = 1 3 (a2 - 1), 即 a1 + a2 = 1 3 (a2 - 1),得 a2 = 1 4 . (2)当 n≥2 时,an = Sn - Sn - 1 = 1 3 (an - 1) - 1 3 (an - 1 - 1), 得 an an - 1 = - 1 2 ,又 a1 = - 1 2 , 所以{an }是首项为 - 1 2 ,公比为 - 1 2 的等比数列. 　 　 典例试做 4:设该等比数列的公比为 q,首项为 a1 ,∵ a2 - a5 = 42,∴ q≠1, 由已知,得 a1 + a1 q + a1 q 2 = 168 a1 q - a1 q 4 = 42{ ,∴ a1 (1 + q + q 2 ) = 168　 ① a1 q(1 - q 3 ) = 42　 ②{ ∵ 1 - q3 = (1 - q)(1 + q + q2 ),∴ 由 ② ① 得 q(1 - q) = 1 4 , ∴ q = 1 2 ,∴ a1 = 42 1 2 - ( 1 2 )4 = 96. 令 G 是 a5 、a7 的等比中项,则应有 G 2 = a5 a7 = a1 q 4 ·a1 q 6 = a21 q 10 = 962 × ( 1 2 )10 = 9, ∴ a5 、a7 的等比中项是 ± 3. 　 　 典例试做 5:(1) 从第一年起,每年车的价值( 万元) 依次设为:a1 , a2 ,a3 ,…,an , 由题 意, 得 a1 = 13. 5, a2 = 13. 5 ( 1 - 10% ), a3 = 13. 5 ( 1 - 10% )2 ,…. 由等比数列定义知数列{an }是等比数列,首项 a1 = 13. 5,公比 q = 1 - 10% = 0. 9, ∴ an = a1 ·q n - 1 = 13. 5 × (0. 9) n - 1 . ∴ 第 n 年车的价值为 an = 13. 5 × (0. 9) n - 1 万元. (2)当他用满 4 年时,车的价值为 a5 = 13. 5 × (0. 9) 5 - 1 = 8. 857. ∴ 用满 4 年卖掉时,他大概能得 8. 857 万元. 课堂达标验收 1. A　 设等比数列的公比为 q, ∵ a1 + a2 = 3,a2 + a3 = q(a1 + a2 ) = 6,∴ q = 2. 又 a1 + a2 = a1 + a1 q = 3,∴ 3a1 = 3. ∴ a1 = 1, ∴ a7 = 2 6 = 64. 2. B　 ∵ 在等比数列{an }中,a3 + a4 = 4,a2 = 2,∴ a3 + a4 = a2 q + a2 q 2 = 2q + 2q2 = 4,即 q2 + q - 2 = 0. 解得 q = 1 或 q = - 2. 故选 B. 3. 4　 由 an = a1 q n - 1 ,得 1 3 = 9 8 × ( 2 3 ) n - 1 ,即( 2 3 ) n - 1 = 8 27 , 故 n = 4. 4. (1)∵ a1 = - 1,an = 3an - 1 - 2n + 3, ∴ a2 = 3a1 - 2 × 2 + 3 = - 4,∴ a3 = 3a2 - 2 × 3 + 3 = - 15. an +1 - (n +1) an -n = 3an -2(n +1) +3 - (n +1) an -n = 3an - 3n an - n = 3(n = 1,2,3,…). 又 a1 - 1 = - 2,∴ {an - n}是以 - 2 为首项,以 3 为公比的等比数列. (2)由(1)知 an - n = - 2·3 n - 1 , 故 an = n - 2·3 n - 1 . 第 2 课时　 等比数列的性质 新知导学 　 　 1. (1)qn - m 　 (2)ap ·aq