2.5.2 数列求和（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）　 ∴ S5 = a1 (1 - q 5 ) 1 - q = 16[1 - ( 1 2 )5 ] 1 2 = 31. (2)32　 当 q = 1 时,显然不符合题意; 当 q≠1 时, a1 (1 - q 3 ) 1 - q = 7 4 , a1 (1 - q 6 ) 1 - q = 63 4 ,{ 解得 a1 = 1 4 , q = 2, { 则 a8 = 14 × 27 = 32. 　 　 跟踪练习 1:(1) 设{an } 的公差为 d,则由已知条件得 a1 + 2d = 2, 3a1 + 3 × 2 2 d = 9 2 ,化简得 a1 + 2d = 2,a1 + d = 3 2 ,解得 a1 = 1,d = 1 2 ,故 通项公式 an = 1 + n - 1 2 ,即 an = n + 1 2 . 　 　 (2)由(1)得 b1 = 1,b4 = a15 = 15 + 1 2 = 8. 设{bn }的公比为 q,则 q 3 = b4 b1 = 8, 从 而 q = 2. 故 { bn } 的 前 n 项 和 Tn = b1(1 -q n) 1 -q = 1 × (1 -2 n) 1 -2 =2n - 1. 　 　 典例试做 2:(1)B　 S4 ,S8 - S4 ,S12 - S8 构成等比数列, 所以(S8 - S4 ) 2 = S4 ·(S12 - S8 ), 因为 S4 = 10,S12 = 130, ∴ (S8 - 10) 2 = 10(130 - S8 ). 解得 S8 = 40. 故选 B. (2) 9 8 　 因为等比数列{an }各项为正,a3 ,a5 , - a4 成等差数列,所 以 a1 q 2 - a1 q 3 = 2a1 q 4 ,2q2 + q - 1 = 0,q = 1 2 或 q = - 1 ( 舍去), S6 S3 = S3 + q 3 S3 S3 = 1 + ( 1 2 )3 = 9 8 . 　 　 跟踪练习 2:(1)A　 S10 ,S20 - S10 ,S30 - S20 成等比数列,所以(S20 - S10 ) 2 = S10 (S30 - S20 ). 即(S20 - 10) 2 = 10(70 - S20 ), 故 S20 = - 20 或 S20 = 30. 又 S20 > 0,所以 S20 = 30,S20 - S10 = 20, S30 - S20 = 40, 故 S40 - S30 = 80,S40 = 150. (2)B　 ∵ a2 ·an - 1 = a1 ·an = 64, 又 a1 + an = 34,可解得 a1 = 2, an = 32 或 a1 = 32,an = 2(舍) Sn = 2(1 - qn ) 1 - q = 2 - 32q 1 - q = 62, 解得 q = 2,∴ an = 2 × 2 n - 1 = 2n = 32,n = 5. 故选 B. 　 　 典例试做 3:(1)设等差数列{an }的公差为 d. 因为 a2 + a4 = 10,所以 2a1 + 4d = 10, 解得 d = 2,所以 an = 2n - 1. (2)设等比数列{bn }的公比为 q, 因为 b2 b4 = a5 ,所以 b1 qb1 q 3 = 9, 解得 q2 = 3. 所以 b2n - 1 = b1 q 2n - 2 = 3n - 1 . 从而 b1 + b3 + b5 + … + b2n - 1 = 1 + 3 + 3 2 + … + 3n - 1 = 3 n - 1 2 . 　 　 跟踪练习 3:(1)由题意,得 an + 1 = 2an + k,bn = an + 1 - an , ∴ bn = 2an + k - an = an + k, ∴ bn + 1 = an + 1 + k = (2an + k) + k = 2(an + k). 即 bn + 1 = 2bn. ∵ b1 ≠0,∴ bn + 1 bn = 2(n∈N∗ ), ∴ 数列{bn }是以 2 为公比的等比数列. (2)由(1),得 bn = an + k 及{bn }是公比为 2 的等比数列,得 Tn = b1 (1 - 2 n ) 1 - 2 = b1 (2 n - 1), 由 bn = an + k 得 Tn = Sn + nk,∴ Sn = b1 (2 n - 1) - nk. ∵ S6 = T4 ,S5 = - 9, ∴ 63b1 - 6k = 15b1 , 31b1 - 5k = - 9, { 解得 k = 8. 　 　 典例试做 4:若 q = 1,则 S3 = 3a1 = 6,符合题意. 此时,q = 1,a3 = a1 =