伤城文章网 > 数学 > 与名师对话2019届高三数学(文)一轮复习课时跟踪训练:第六章 数列 课时跟踪训练33 Word版含解析

与名师对话2019届高三数学(文)一轮复习课时跟踪训练:第六章 数列 课时跟踪训练33 Word版含解析


课时跟踪训练(三十三) [基础巩固] 一、选择题 1.(2018· 湖南师大附中月考)已知公差不为 0 的等差数列{an}满 S3-S2 足 a1,a3,a4 成等比数列,Sn 为数列{an}的前 n 项和,则 的值 S5-S3 为( ) A.2 B.3 C.-2 D.-3 [解析] 设等差数列的公差为 d,首项为 a1,所以 a3=a1+2d, a4=a1+3d. 因为 a1、a3、a4 成等比数列, 所以(a1+2d)2=a1(a1+3d),解得:a1=-4d. S3-S2 a1+2d 所以 = =2,故选 A. S5-S3 2a1+7d [答案] A 2.(2017· 河南百校联盟质量监测)已知等差数列{an}的前 n 项和 为 Sn,S5=-20,则-6a4+3a5=( A.-20 B.4 C.12 D.20 5?a1+a5? [解析] 设{an}的公差为 d,∵S5= =-20,∴a1+a5=- 2 8,∴a3=-4.又-6a4+3a5=-6(a3+d)+3(a3+2d)=-3a3=12.选 C. [答案] C 3.已知等比数列{an}的首项为 1,若 4a1,2a2,a3 成等差数列,则 ) ?1? 数列?a ?的前 5 项和为( ? n? ) 31 33 16 A.16 B.2 C.16 D.33 [解析] 设数列{an}的公比为 q,则有 4+q2=2×2q,解得 q=2, 所以 an=2n-1. ?1? 1-?2?5 1 1 31 ? ? = ,所以 S = = 5 an 2n-1 1 16.故选 A. 1-2 [答案] A 4.已知数列{an}是等差数列,a1=tan225° ,a5=13a1,设 Sn 为数 列{(-1)nan}的前 n 项和,则 S2018=( ) A.2018 B.-2018 C.3027 D.-3027 [解析] 由题意得 a1=1,a5=13,∵{an}是等差数列,∴公差 d =3, ∴an=3n-2, ∴S2018=-1+4-7+10-13+17+…-6049+6052 2018 =3× 2 =3027,选 C. [答案] C 5.(2017· 安徽安庆模拟)已知数列{an}满足 an+2=-an(n∈N+), 且 a1=1,a2=2,则数列{an}的前 2017 项的和为( A.2 B.-3 C.3 D.1 [解析] ∵an+2=-an=-(-an-2)=an-2,n>2,∴数列{an}是以 4 为周期的周期数列.S2017=504(a1+a2+a3+a4)+a2017=504(a1+a2- a1-a2)+a504×4+1=a1=1.故选 D. [答案] D ) 1 1 1 1 6. 2 + 2 + 2 +…+ 的值为( 2 -1 3 -1 4 -1 ?n+1?2-1 n+1 A. 2?n+2? n+1 3 B.4- 2?n+2? ) 1 ? 3 1? 1 3 1 1 C.4-2?n+1+n+2? D.2- - n+1 n+2 ? ? 1 1 ? 1 1 1 1? ? - ? [解析] 因为 = = = ?n n+2? 2 2 2 ?n+1? -1 n +2n n?n+2? ? ? 所以原式= 1? ?1 1? ?1 1? 1?? ??1- ?+? - ?+? - ?+…+ 3? ?2 4? ?3 5? 2?? ?1 1 ?? 1? 1 1 1 ? ? - ?? ?1+ - ? - = 2 n+1 n+2? ?n n+2?? 2? ? ?? ? ? 3 1? 1 + 1 ? ? =4-2? ?n+1 n+2?,故选 C. ? ? [答案] C 二、填空题 7.若数列{an}的通项公式为 an= 则 S16=________. [解析] 由 an= 1 n+ n+2 1 =2? ? n+2- n? ?, 1 n+ n+2 ,前 n 项和为 Sn, 1 得 S16=2( 3-1+ 4- 2+ 5- 3+…+ 17- 15+ 18- 17+2 2-1 1 16)=2( 18+ 17- 2-1)= . 2 [答案] 17+2 2-1 2 1 8.数列{an}满足 an+an+1=2(n∈N*),且 a1=1,Sn 是数列{an} 的前 n 项和,则 S21=________. 1 [解析] 依题意得 an+an+1=an+1+an+2=2,则 an+2=an,即数 列{an}中的奇数项、偶数项分别相等,则 a21=a1=1,S21=(a1+a2) 1 +(a3+a4)+…+(a19+a20)+a21=10(a1+a2)+a21=10×2+1=6. [答案] 6 9. (2017· 陕西西安期中)如果数列{an}的前 n 项之和为 Sn=3+2n, 2 2 2 那么 a2 1+a2+a3+…+an=________. [解析] ∵Sn=3+2n,∴Sn-1=3+2n-1(n≥2),∴an=2n-2n-1=2n -1,∴a2=4n-1,n=1 n 时 a1=S1=5, 4?1-4n-1? 4n+71 = 3 ; 1-4 4n+71 3 . ∴当 n≥2 2 2 2 时,a1 +a2 2+a3+…+an=25+ 当 n=1 时 a2 1=25 也适合上式,故 2 a1 +…+a2 n= 4n+71 [答案] 3 三、解答题 10.(2017· 全国卷Ⅲ)设数列{an}满足 a1+3a2+…+(2n-1)an= 2n. (1)求{an}的通项公式; ? an ? (2)求数列?2n+1?的前 n 项和. ? ? [解] (1)因为 a1+3a2+…+(2n-1)an=2n,故当 n≥2 时, a1+3a2+…+(2n-3)an-1=2(n-1). 2 两式相减得(2n-1)an=2,所以 an= (n≥2). 2n-1 2 又由题设可得 a1=2 也适合,从而{an}的通项公式为 an= . 2n-1 ? ? an

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