黄骅贴吧百度贴吧:2019年《南方新课堂·高考总复习》数学(理科)作业及测试:课时作业 专题三数列与不等式 Word版含解析

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专题三 数列与不等式 Sn 7n+45 an 1.已知等差数列{an},{bn}的前 n 项和分别为 Sn,Tn,且 = ,则使得 为整数 Tn n-3 bn 的正整数 n 的个数是( ) A.3 B.4 C.5 D.6 2.已知等差数列{an}的公差 d≠0,且 a1,a3,a13 成等比数列,若 a1=1,Sn 为数列{an} 2Sn+16 的前 n 项和,则 的最小值为( ) an+3 9 A.4 B.3 C.2 3-2 D. 2 3.(2015 年新课标Ⅱ)设 Sn 是数列{an}的前 n 项和,且 a1=-1,an+1=SnSn+1,则 Sn= ________. 4.设数列{an}的前 n 项和为 Sn,且满足 an+Sn=1,则 Sn 的取值范围是( ) A.(0,1) B.(0,+∞) 1 ? ?1 ? C.? ?2,1? D.?2,+∞? 5.(2017 年广东调研)设 Rn 是等比数列{an}的前 n 项的积,若 25(a1+a3)=1,a5=27a2, 则当 Rn 取最小值时,n=______. 6.(2017 年新课标Ⅰ)几名大学生响应国家的创业号召,开发了一款应用软件.为激发 大家学习数学的兴趣, 他们推出了“解数学题获取软件激活码”的活动. 这款软件的激活码 为下面数学问题的答案:已知数列 1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,…,其中第一项是 20,接下 来的两项是 20,21,再接下来的三项是 20,21,22,依此类推.求满足如下条件的最小整数 N: N>100 且该数列的前 N 项和为 2 的整数幂.那么该款软件的激活码是( ) A.440 B.330 C.220 D.110 7.(2016 年新课标Ⅲ)已知各项都为正数的数列{an}满足 a1=1,a2 n-(2an+1-1)an-2an+ 1=0. (1)求 a2,a3; (2)求{an}的通项公式. 8.(2017 年广东揭阳一模)设等差数列{an}的前 n 项和为 Sn,且 S4=4S2,a2n=2an+1- 3. (1)求数列{an}的通项公式; 2n+3 (2)设数列{bn}满足 a1b1+a2b2+…+anbn=3- n ,求{bn}的前 n 项和 Tn. 2 9.(2017 年广东汕头一模)已知数列{an}的前 n 项和为 Sn,a1=2,an+1=Sn+2. (1)求数列{an}的通项公式; ? 1 ? (2)已知 bn=log2an,求数列?b b ?的前 n 项和 Tn. ? n n+1? 10.(2017 年天津)已知{an}为等差数列,前 n 项和为 Sn(n∈N*),{bn}是首项为 2 的等比 数列,且公比大于 0,b2+b3=12,b3=a4-2a1,S11=11b4. (1)求{an}和{bn}的通项公式; (2)求数列{a2nb2n-1}的前 n 项和(n∈N*). 专题三 1.B 数列与不等式 an 2an a1+a2n-1 S2n-1 14n+38 7n+19 33 解析: = = = = = =7+ ,∴n-2=-1 bn 2bn b1+b2n-1 T2n-1 2n-4 n-2 n-2 Sn Tn+45 或 1 或 3 或 11 或 33,∴n=1 或 3 或 5 或 13 或 35.当 n=3 时, = 中分母为零,所以 Tn n-3 舍去. 2 2.A 解析:由 a1,a3,a13 成等比数列,得 a3 =a1a13?(a1+2d)2=a1(a1+12d)?4d2= 2Sn+16 n2+8 9 8a1d.因为 d≠0,因此 d=2a1=2,Sn=n ,an=2n-1,从而 = =(n+1)+ - an+3 n+1 n+1 2 2≥2 9 ?n+1?× -2=4,当且仅当 n=2 时取等号.故选 A. n+1 1 1 3.- 解析:由已知,得 an+1=Sn+1-Sn=Sn+1· Sn,两边同时除以-Sn+1· Sn,得 n Sn+1 1 ?1? 1 - =-1,故数列?S ?是以-1 为首项,-1 为公差的等差数列.则 =-1-(n-1)=-n. Sn Sn ? n? 1 所以 Sn=- . n 1 4.C 解析:当 n=1 时,a1= .当 n≥2 时,an-1+Sn-1=1,得 an-an-1+an=0,即 2 2an=an-1. 1 1 ∴数列{an}是首项为 ,公比为 的等比数列. 2 2 1 ? ?1?n? × 1- 2 ? ?2? ? 1?n ∴Sn= =1-? ?2? . 1 1- 2 1 ? ∴Sn∈? ?2,1?. a5 5.6 解析:设公比为 q,则 q3= =27.所以 q=3.由 25(a1+a3)=1,得 25(a1+a1· 32) a2 ? ? ? 250 ≤1, 3 ?an≤1, 1 =1,解得 a1= .则 an= .则要使 Rn 取得最小值,必有? 即? n 250 250 ? ?an+1>1, ? 3 >1, n-1 3n-1 所 ?250 以 250<3n≤750,解得 n=6. 6.A 解析:由题意,得数列如下: 1, 1,2, 1,2,4, … 则该数列的前 1+2+…+k= k?k+1? 项和为 2 S? ?k?k+1?? ?=1+(1+2)+…+(1+2+…+2k)=2k+1-k-2. 2 ? ? k?k+1? 要使 >100,有 k≥14,此时 k+2<2k+1.所以 k+2 是之后的等比数列 1,2,…,2k 2 +1 的部分和,即 k+2=1+2+…+2t-1=2t-1. 所以 k=2t-3≥14,则 t≥5,此时 k=25-3=29. 29×30 对应满足的最小条件为 N= +5=440.故选 A. 2 1 1 7.解:(1)由题意,得 a2= ,a3= . 2 4 (2)由 a2 n-(2an+1-1)an-2an+1=0,得 2an+1(an+1)

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