2.3.2 等差数列习题课（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）　 (2)1 008　 由等差数列的性质知,S2 017 = 2 017a1 009 < 0,所以 a1 009 < 0,又 S2 016 = 2 016(a1 + a2 016 ) 2 = 1 008( a1 008 + a1 009 ) > 0,所以 a1 008 + a1 009 > 0,而 a1 009 < 0,故 a1 008 > 0. 因此当 n = 1 008 时,Sn 最大. 　 　 跟踪练习 3:7　 S3 = S11 ,所以其对称轴为 n = 3 + 11 2 = 7,知 n = 7 时 Sn 取最大值. 　 　 典例试做 4:a1 = S1 = 6, n≥2 时,an = Sn - Sn - 1 = (n 2 + 3n + 2) - [(n - 1)2 + 3(n - 1) + 2] = 2n + 2, ∴ an = 6　 　 　 (n = 1) 2n + 2 (n≥2){ ,显然 a2 - a1 = 4 - 6 = - 2,a3 - a2 = 2, ∴ {an }不是等差数列. 　 　 典例试做 5:(1)当 n = 1 时,a1 = S1 = 3 1 - 2 = 1;当 n≥2 时,Sn - 1 = 3n - 1 - 2, 则 an = Sn - Sn - 1 = (3 n - 2) - (3n - 1 - 2) = 3n - 3n - 1 = 3·3n - 1 - 3n - 1 = 2·3n - 1 . 此 时 若 n = 1, 则 2 · 3n - 1 = 2 · 31 - 1 = 2 ≠ a1 , 故 an = 1(n = 1) 2·3n - 1 (n≥2){ . (2)当 n = 1 时,a1 = S1 = 1 4 (a1 + 1) 2 ,解得 a1 = 1. 当 n≥2 时,an = Sn - Sn - 1 = 1 4 [(an + 1) 2 - ( an - 1 + 1) 2 ],整理得 (an + an - 1 )(an - an - 1 - 2) = 0. ∵ an + an - 1 ≠0,∴ an - an - 1 = 2. ∴ {an } 是以 a1 = 1 为首项,2 为公 差的等差数列,∴ an = 2n - 1. 课堂达标验收 1. B　 S11 = 11(a1 + a11 ) 2 = 11(a4 + a8 ) 2 = 11 × 16 2 = 88. 2. C　 由题意得 a1 + 2d = 6 3a1 + 1 2 × 3 × 2d = 12{ , 解得 a1 = 2 d = 2{ . 3. C　 am = Sm - Sm - 1 = 2,am + 1 = Sm + 1 - Sm = 3,公差 d = am + 1 - am = 3 - 2 = 1. 由 Sm = m(a1 + am ) 2 = 0,得 a1 = - am = - 2. ∴ am = - 2 + (m - 1)·1 = 2,解得 m = 5. 4. - 3　 已知等差数列{an } 的前 5 项和 S5 = 25,所以 S5 = 5(a1 + a5 ) 2 = 5a3 = 25,解得 a3 = 5. 已知 a4 = 3,则公差 d = a4 - a3 = - 2. 所以 a7 = a3 + 4d = 5 - 8 = - 3. 5. (1)设公差为 d,由题意得 a1 + 4d = 11 a1 + 7d = 5 { , 解得 a1 = 19 d = - 2{ . ∴ an = a1 + (n - 1)d = 19 - 2(n - 1) = 21 - 2n. (2)由题意,得 a1 + d + a1 + 3d = 4 a1 + 2d + a1 + 4d = 10 { , ∴ a1 + 2d = 2 a1 + 3d = 5 { , 解得 a1 = - 4 d = 3{ . ∴ S10 = ( - 4) × 10 + 10 × 9 2 × 3 = 95. 第 2 课时　 等差数列习题课 新知导学 　 　 1. Sn - Sn - 1 　 an = Sn ,n = 1 Sn - Sn - 1 ,n≥2 { 　 2. 1anbn{ }　 1 d ( 1 an - 1 bn ) 预习自测 1. A　 a8 + a9 + a10 + a11 + a12 = S12 - S7 = 122 + 12 + 1 - 72 - 7 - 1 = 100. 2. B　 an = Sn - Sn - 1 = 2 n - 2n - 1 = 2n - 1 , 又 S1 = 2 1 = 2,a1 = 2 1 - 1 = 1. 不符. ∴ an = 2,n = 1 2n - 1 ,n > 1{ ∴ a8 = 2