3.1.1 不等关系与不等式的性质（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）

现行旧教材·高中新课程学习指导 = n(n + 1) + 1 2 - 1 2n + 1 . 　 　 典例试做 8:∵ 1 n2 - 1 = 1 (n - 1)(n + 1) = 1 2 ( 1 n - 1 - 1 n + 1 ), ∴ 1 22 - 1 + 1 32 - 1 + 1 42 - 1 + … + 1 n2 - 1 = 1 2 [(1 - 1 3 ) + ( 1 2 - 1 4 ) + ( 1 3 - 1 5 ) + … + ( 1 n - 1 - 1 n + 1 )] = 1 2 (1 + 1 2 - 1 n - 1 n + 1 ) = 3 4 - 2n + 1 2n(n + 1) (n≥2). 　 　 典例试做 9:Sn = 1 + 2x + 3x 2 + 4x3 + … + nxn - 1 ,① xSn = x + 2x 2 + 3x3 + … + (n - 1)xn - 1 + nxn ,② 由① - ②,得(1 - x)Sn = 1 + x + x 2 + … + xn - 1 - nxn , 当 x≠1 时,(1 - x)Sn = 1 - xn 1 - x - nxn = 1 - x n - nxn + nxn + 1 1 - x = 1 - (1 + n)x n + nxn + 1 1 - x , ∴ Sn = 1 - (1 + n)xn + nxn + 1 (1 - x)2 ; 当 x = 1 时,Sn = 1 + 2 + 3 + 4 + … + n = n(1 + n) 2 , ∴ Sn = n(1 + n) 2 (x = 1) 1 - (1 + n)xn + nxn + 1 (1 - x)2 (x≠1){ . 　 　 典例试做 10:(1)证明:由题设得 4( an + 1 + bn + 1 ) = 2( an + bn ),即 an + 1 + bn + 1 = 1 2 (an + bn ). 又因为 a1 + b1 = 1, 所以{an + bn }是首项为 1,公比为 1 2 的等比数列. 由题设得 4(an + 1 - bn + 1 ) = 4(an - bn ) + 8, 即 an + 1 - bn + 1 = an - bn + 2. 又因为 a1 - b1 = 1, 所以{an - bn }是首项为 1,公差为 2 的等差数列. (2)解:由(1)知,an + bn = 1 2n - 1 ,an - bn = 2n - 1, 所以 an = 1 2 [(an + bn ) + (an - bn )] = 1 2n + n - 1 2 , bn = 1 2 [(an + bn ) - (an - bn )] = 1 2n - n + 1 2 . 　 　 典例试做 11:(1)当 n = 1 时,S1 = a(S1 - a1 + 1), 所以 a1 = a, 当 n≥2 时,Sn = a(Sn - an + 1),① Sn - 1 = a(Sn - 1 - an - 1 + 1),② 由① - ②,得 an = a·an - 1 ,即 an an - 1 = a, 所以{an }是首项 a1 = a,公比为 a 的等比数列,所以 an = a·a n - 1 = an. a2 = a 2 ,a3 = a 3 . 由 4a3 是 a1 与 2a2 的等差中项, 可得 8a3 = a1 + 2a2 , 即 8a 3 = a + 2a2 , 因为 a≠0,整理得 8a2 - 2a - 1 = 0, 即(2a - 1)(4a + 1) = 0,解得 a = 1 2 或 a = - 1 4 ( 舍去),所以 an = ( 1 2 ) n = 1 2n . (2)由(1),得 bn = 2n + 1 an = (2n + 1)·2n , 所以 Tn = 3 × 2 + 5 × 2 2 + 7 × 23 + … + (2n - 1) ·2n - 1 + (2n + 1) · 2n ,① 2Tn = 3 × 2 2 + 5 × 23 + 7 × 24 + … + (2n - 1) ·2n + (2n + 1) · 2n + 1 ,② 由① - ②,得 - Tn = 3 × 2 + 2(2 2 + 23 + … + 2n ) - (2n + 1)·2n + 1 = 6 + 2 × 2 2 - 2n + 1 1 - 2 - (2n + 1)·2n + 1 = - 2 + 2n + 2 - (2n + 1)·2n + 1 = - 2 - (2n - 1)·2n + 1 , 所以 Tn = 2 + (2n - 1)·2 n + 1 . 　 　 典例试做 12:(1)因为 an + 1 = 2Sn + 1,n