2.2.1 等差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 综上所述,数列{bn }是递增数列时 a 的取值范围是(0, 1 2 )∪(1, + ∞ ). 　 　 典例试做 5:(1)an = an - 1 + 1 n(n - 1) ,(n≥2), ∴ a2 - a1 = 1 2 × 1 ,a3 - a2 = 1 3 × 2 ,a4 - a3 = 1 4 × 3 ,……,an - an - 1 = 1 n(n - 1) (n≥2). 以上各式相加,得 an - a1 = 1 2 × 1 + 1 3 × 2 + 1 4 × 3 + … + 1 n(n - 1) . 又 1 n(n - 1) = 1 n - 1 - 1 n , ∴ an - a1 = 1 - 1 2 + 1 2 - 1 3 + 1 3 - 1 4 + … + 1 n - 1 - 1 n = 1 - 1 n , ∴ an = a1 + 1 - 1 n = 2 - 1 n = 2n - 1 n (n≥2). a1 = 1 满足上式, ∴ an = 2n - 1 n (n∈N∗ ). (2)∵ (n + 1)an + 1 = nan ,∴ an + 1 an = n n + 1 . ∴ a2 a1 = 1 2 , a3 a2 = 2 3 , a4 a3 = 3 4 ,……, an an - 1 = n - 1 n (n≥2). 以上各式相乘,得 an a1 = 1 n . ∵ an = a1 n = 1 n ( n≥2),又 a1 = 1 满足 上式,∴ an = 1 n (n∈N∗ ). 课堂达标验收 1. A　 ∵ f(n + 1) - f(n) = 3(n∈N∗ ), ∴ f(2) > f(1),f(3) > f(2),f(4) > f(3),…, f(n + 1) > f(n),…, ∴ f(n)是递增数列. 2. A　 n = 3 时,a3 = a2 + 1 a1 = 3 + 1 = 4; n = 4 时,a4 = a3 + 1 a2 = 4 + 1 3 = 13 3 ; n = 5 时,a5 = a4 + 1 a3 = 13 3 + 1 4 = 55 12 . 故选 A. 3. B　 ∵ a1 = - 2,an + 1 = 1 - 1 an , ∴ a2 = 1 + 1 2 = 3 2 ,a3 = 1 - 1 a2 = 1 - 2 3 = 1 3 , a4 = 1 - 1 a3 = 1 - 3 = - 2, ∴ 数列{an }是周期 T = 3 的周期数列, ∴ a2 019 = a3 = 1 3 . 4. 6 7 　 解法一:∵ a1 = 1,an ·an - 1 = n - 1(n≥2), ∴ a2 = 1 a1 = 1,a3 = 2 a2 = 2, a4 = 3 a3 = 3 2 ,a5 = 4 a4 = 8 3 , a6 = 5 a5 = 15 8 ,a7 = 6 a6 = 16 5 ,a8 = 7 a7 = 35 16 . ∴ a6 a8 = 15 8 · 16 35 = 6 7 . 解法二:由题意得,a7 ·a6 = 6,a8 ·a7 = 7,两式相除即得 a6 a8 = 6 7 . 5. ∵ a1 = 2,an + 1 = an + n, ∴ 当 n = 1 时,a2 = a1 + 1 = 2 + 1 = 3; 当 n = 2 时,a3 = a2 + 2 = 3 + 2 = 5; 当 n = 3 时,a4 = a3 + 3 = 5 + 3 = 8; 当 n = 4 时,a5 = a4 + 4 = 8 + 4 = 12,即 a5 = 12. 2. 2　 等差数列 第 1 课时　 等差数列的概念与通项公式 新知导学 　 　 1. 第 2 项　 同一个常数　 公差　 常数列 2. a1 + (n - 1)d 3. 等差中项　 a + b 2 　 预习自测 1. (1) × 　 如数列 2,7,9,1. 虽然 7 - 2 = 5,9 - 7 = 2,1 - 9 = - 8,每一项 与前一项的差都是常数,但不是同一个常数,故不是等差数列. (2)√　 因为从第 2 项起每一项与前一项的差是同一个常数 0. (3)√　 符合等差数列的定义,从第二项起每一项与它的前一项的差 是 1. (4) × 　 因为 m 是 2 与 14 的等差中项,所以 2m = 2 + 14,则 m = 8. 2. - 2　 d = an - an - 1 = 3 - 2n - 3 + 2(n - 1) = - 2. 3. 3　 设方程 x2 - 6x + 1 = 0 的两根为 x1 ,x2 ,则 x1 + x2 = 6. 所以其等差 中项为 x1 + x2 2 = 3. 4. 17　 因为在等差数列{a