mg5349.com:高二数学人教A必修5练习:2.3.1 等差数列的前n项和 Word版含解析

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课时训练 9 等差数列的前 n 项和 一、等差数列前 n 项和公式及应用 1.在等差数列{an}中,d=2,an=11,Sn=35,则 a1 为( A.5 或 7 答案:D 解析:a1+(n-1)×2=11 ①,Sn=na1+ (-1) ×2=35 2 ) D.3 或-1 B.3 或 5 C.7 或-1 ②,由①②解得 a1=3 或 a1=-1. 经检验,a1=3 与 a1=-1 均符合题意,故选 D. 2.已知等差数列{an}的前 n 项和为 Sn,若 a4=18-a5,则 S8 等于( A.18 答案:D 解析:∵a4=18-a5,∴a4+a5=18. B.36 C.54 D.72 ) ∴S8= 8(1 +8 ) =4(a1+a8)=4(a4+a5)=72. 2 3.(2015 河北邯郸三校联考,2)等差数列{an}中,a1+a2+a3=-24,a18+a19+a20=78,则此数列前 20 项和等于 ( ) B.180 C.200 D.220 A.160 答案:B 解析:∵a1+a2+a3=-24,a18+a19+a20=78, ∴a1+a20+a2+a19+a3+a18=54=3(a1+a20). ∴a1+a20=18.∴S20= 20(1 +20 ) =180.故选 B. 2 15 4.设 Sn 为公差不为零的等差数列{an}的前 n 项和.若 S9=3a8,则3 =( 5 ) A.15 答案:A 解析:由 S9=3a8,得 B.17 C.19 D.21 9(1 +9 ) 2 15 15 = 2(a1+a15),即 9a5= 5 ,所以3 =15. 5 3 5.有一个凸 n 边形,各内角的度数成等差数列,公差是 10°,最小角为 100°,则边数 n= 答案:8 . 解析:n×100°+ 2 ×10°=(n-2)×180°,解得 n=8 或 n=9.又 an=100°+(n-1)×10°<180°,∴n=8. 6.已知等差数列{an}的前三项为 a-1,4,2a,记前 n 项和为 Sn.设 Sk=2 550,求 a 和 k 的值. 解:设{an}的公差为 d,由已知得,a1=a-1,a2=4,a3=2a. 又 2a2=a1+a3,∴8=(a-1)+2a,∴a=3, (-1) ∴a1=2,d=a2-a1=2. 由 Sk=ka1+ (-1) d,得 2 2k+ (-1) ×2=2550, 2 即 k2+k-2550=0,解得 k=50 或 k=-51(舍去),∴a=3,k=50. 二、由 Sn 求解数列的通项公式 7.设数列{an}的前 n 项和 Sn=n2+1,则数列{an}的通项公式为 答案:an= 2, = 1, 2-1, ≥ 2 . 解析:当 n≥2 时,an=Sn-Sn-1=n2+1-(n-1)2-1=2n-1. 当 n=1 时,a1=S1=1+1=2 不适合上式. 2, = 1, 2-1, ≥ 2. ∴数列{an}的通项公式为 an= 8.已知数列{an}的前 n 项和为 Sn=3n-2,求数列{an}的通项公式. 解:当 n=1 时,a1=S1=31-2=1; 当 n≥2 时,an=Sn-Sn-1=(3n-2)-(3n-1-2)=2×3n-1,而 2×31-1=2≠1. 故数列{an}的通项公式为 an= 1, = 1, 2 × 3 -1 , ≥ 2. 9.已知数列{an}的前 n 项和为 Sn,且满足 a1=1,an+2SnSn-1=0(n≥2). (1)求证:数列 是等差数列; (2)求{an}的通项公式. (1)证明:∵n≥2 时,an=Sn-Sn-1, 又 an+2SnSn-1=0,∴Sn-Sn-1+2SnSn-1=0. 1 ∵Sn≠0,两边同除以 SnSn-1,得 1 -1 ? +2=0,即 ? =2(n≥2), -1 1 1 1 1 1 ∴数列 是等差数列. 1 (2)解:∵a1=1, = =1, 1 1 ∴ =1+(n-1)×2=2n-1,∴Sn= 1 1 . 2-1 当 n≥2 时,an=Sn-Sn-1= =2 . (2-1)(2-3) 1 1 ? 2-1 2(-1)-1 而- 2 =2≠1,故{an}的通项公式 an= (2×1-1)(2×1-3) 1, = 1, 2 , (2-1)(2-3) ≥ 2. (建议用时:30 分钟) 1.在等差数列{an}中,若 a1-a4+a8-a12+a15=2,则 S15 等于( A.28 答案:B 解析:∵a1-a4+a8-a12+a15 =(a1+a15)-(a4+a12)+a8=a8=2. B.30 C.31 ) D.32 ∴S15= 15(1 +15 ) 2 = 15×28 =30. 2 2.在等差数列{an}中,公差 d≠0,首项 a1≠d.如果这个数列的前 20 项的和 S20=10M,则 M 应是( A.a5+a15 C.2a1+19d 答案:C 解析:∵S20=20a1+ 2 d=10(2a1+19d)=10M, 20×19 ) B.a2+2a10 D.a20+d ∴M=2a1+19d. 3.已知数列{an}为等差数列,其前 n 项的和为 Sn,若 a3=6,S3=12,则公差 d=( A.1 答案:B 解析:在等差数列中,S3= 3(1 +3 ) 2 ) B.2 C.3 D. 5 3 = 3(1 +6) =12,解得 a1=2,所以解得 2 d=2,选 B. 4.将含有 k 项的等差数列插入 4 和 67 之间,结果仍成一新的等差数列,并且新的等差数列所有项的和 是 781,则 k 的值为( A.20 答案:A 解析:由数列前 n 项和公式可得 (+2)(4+67) =781,解得 2 1 ) B.21 C.22 D.24 k=20. ) 5.(20

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