2.4.1 等比数列的概念与通项公式（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）

现行旧教材·高中新课程学习指导 　 　 跟踪练习 3:(1)24　 36　 a1 = - 10,d = 2, 所以 an = - 10 + 2(n - 1) = 2n - 12. a6 = 0, 故 S3 = | - 10 | + | - 8 | + | - 6 | = 24, S8 = | a1 | + | a2 | + | a3 | + … | a6 | + | a7 | + | a8 | = - a1 - a2 - … - a6 + a7 + a8 = 36. (2)①因为在等差数列{an }中,a1 + a5 = 8,a4 = 2, 所以 2a1 + 4d = 8, a1 + 3d = 2, { 解得 a1 = 8,d = - 2, 所以 an = 8 + (n - 1) × ( - 2) = 10 - 2n. ②由 an = 10 - 2n≥0,得 n≤5, a5 = 0,a6 = - 2 < 0, 因为 Tn = |a1 | + |a2 | + |a3 | + … + |an |,所以当 n≤5 时, Tn = 8n + n(n - 1) 2 × ( - 2) = 9n - n2 . 当 n > 5 时, Tn = - [8n + n(n - 1) 2 × ( - 2)] + 2(9 × 5 - 52 ) = n2 - 9n + 40. 所以 Tn = 9n - n2 ,n≤5, n2 - 9n + 40,n > 5.{ 　 　 典例试做 4:∵ 1 n(n + 2) = 1 2 ( 1 n - 1 n + 2 ), ∴ 数列{ 1 n(n + 2) }的前 n 项和 Sn = 1 2 (1 - 1 3 + 1 2 - 1 4 + 1 3 - 1 5 + … + 1 n - 1 - 1 n + 1 + 1 n - 1 n + 2 ) = 1 2 (1 + 1 2 - 1 n + 1 - 1 n + 2 ) = 3 4 - 2n + 3 2(n + 1)(n + 2) . 　 　 典例试做 5:(1)∵ 对任意的正整数 n,2 Sn = an + 1①恒成立,当 n = 1 时,2 a1 = a1 + 1,即( a1 - 1) 2 = 0,∴ a1 = 1. 当 n≥2 时,有 2 Sn - 1 = an - 1 + 1. ②,① 2 - ②2 得 4an = a 2 n - a 2 n - 1 + 2an - 2an - 1 , 即(an + an - 1 )(an - an - 1 - 2) = 0. ∵ an > 0,∴ an + an - 1 > 0, ∴ an - an - 1 = 2, ∴ 数列{an }是首项为 1,公差为 2 的等差数列,∴ an = 1 + (n - 1) × 2 = 2n - 1. (2)∵ an = 2n - 1,∴ bn = 1 (2n - 1)(2n + 1) = 1 2 1 2n - 1 - 1 2n + 1( ), ∴ Bn = b1 + b2 + b3 + … + bn = 1 2 1 - 1 3( ) + 1 2 1 3 - 1 5( ) + 1 2 1 5 - 1 7( ) + … + 1 2 1 2 n - 1 - 1 2 n + 1( ) = 1 2 1 - 1 2n + 1( ) = n 2n + 1 . 课堂达标验收 1. D　 ∵ S7 = 7a4 = 42,∴ a4 = 6,∴ d = a7 - a4 7 - 4 = 2 3 ,故选 D. 2. C　 an = 120 + 5(n - 1) = 5n + 115, 由 an < 180 得 n < 13 且 n∈N ∗ , 由 n 边形内角和定理得, (n - 2) × 180 = n × 120 + n(n - 1) 2 × 5. 解得 n = 16 或 n = 9 ∵ n < 13,∴ n = 9. 3. 2 000　 假设 20 位同学是 1 号到 20 号依次排列,使每位同学从各自树 坑出发前来领取树苗往返所走的路程总和最小,则树苗需放在第 10 或第 11 号树坑旁,此时两侧的同学所走的路程分别组成以 20 为首 项,20 为公差的等差数列,故所有同学往返的总路程为 S = 9 × 20 + 9 × 8 2 × 20 + 10 × 20 + 10 × 9 2 × 20 = 2 000. 4. (1){an}为等差数列,a4 +a5 +a6 +a7 +a8 =25, ∵ 5a6 = 25,∴ a6 = 5. ∵ {Sn }为{an }的前 n 项和, ∴ S12 = 12(a1 + a12 ) 2 = 6(a6 + a7 ) = 6(5 + a7 ) = 54, ∴ a7 = 4,∴ d =