课时提升作业(五) 5

6《老子》四章*五石之瓠《〈老子〉四章》时间：45分钟满分：70分得分：______分一、基础素养巩固(24分)1.下列各句中，没有通假字的一项是()(3分)A.死而不亡者寿B.其在道也，曰余食赘行C.起于累土D.其未兆易谋，其脆易泮解析B项，“行”同“形”。

C项，“累”同“蔂”。

D项，“泮”同“判”。

答案A2.对下列加点词的解释，不正确的一项是()(3分)A.生于毫末．．毫末：极其细微的萌芽。

B.复．众人之所过复：重复。

C.故有道者不处．处：为，做。

D.企．者不立企：踮着脚跟。

解析B项，复：弥补、补救。

答案B3.对下列各句中加点词的意义和用法的解释，不正确的一项是()(3分)A.民之．从事，常于几成而败之之：助词，用于主谓之间。

B.三十辐共一毂，当其．无，有车之用其：代词，它。

C.跨者．不行，自见者不明者：助词，……的(人)。

D.合抱之木，生于．毫末于：介词，在。

解析于：介词，从。

答案D4.下列各句中，加点词的古今意义相同的一项是()(3分)A.企者不立，跨者不行．．．． B.民之从事C.苟能充之，足以保．四海D.复众人．．之所过解析A项，“不行”在句中指“行走不稳”。

B项，“从事”在句中指“做事”。

C项，“保”在句中指“安定”。

答案D5.下列各句中加点词的活用类型，与其他三项不同的一项是()(3分)A.常于几成而败．之B.二月草已芽．C.风．乎舞雩D.刑．于寡妻解析A项，使动用法，使……失败。

B、C、D三项都是名词活用作动词，分别为发芽、吹风、做榜样的意思。

答案A6.下列各句中，与例句句式相同的一项是()(3分)例句：治之于未乱A.自知者明B.为之于未有C.蚓无爪牙之利D.外欺于张仪解析B项和例句相同，均为状语后置句。

A项，宾语前置句。

C项，定语后置句。

D项，被动句。

答案B7.补写出下列句子中的空缺部分。

(6分)(1)企者不立，跨者不行，________，自是者不彰，自伐者无功，________。

数列的综合应用(20分钟40分)一、选择题(每小题5分,共25分)1.(2014·北京高考)设{a n}是公比为q的等比数列,则“q>1”是“{a n}为递增数列”的( )A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件【解析】选D.当a1<0,q>1时,{a n}是递减数列;当{a n}为递增数列时,a1<0,0<q<1或a1>0,q>1.因此,“q>1”是“{a n}为递增数列”的既不充分也不必要条件.【加固训练】(2016·南昌模拟)在公差不为0的等差数列{a n}中,2a3-+2a11=0,数列{b n}是等比数列,且b7=a7,则b6b8= ( )A.2B.4C.8D.16【解析】选D.因为{a n}是等差数列,所以a3+a11=2a7,所以2a3-+2a11=4a7-=0,解得a7=0或4,因为{b n}为等比数列,所以b n≠0,所以b7=a7=4,b6b8==16.2.设y=f(x)是一次函数,若f(0)=1,且f(1),f(4),f(13)成等比数列,则f(2)+f(4)+…+f(2n)等于( )A.n(2n+3)B.n(n+4)C.2n(2n+3)D.2n(n+4)【解析】选A.由题意可设f(x)=kx+1(k≠0),则(4k+1)2=(k+1)×(13k+1),解得k=2,f(2)+f(4)+…+f(2n)=(2×2+1)+(2×4+1)+…+(2×2n+1)=2n2+3n=n(2n+3).3.(2016·大同模拟)已知正项等差数列{a n}满足:a n+1+a n-1=(n≥2),等比数列{b n}满足:b n+1b n-1=2b n(n≥2),则log2(a2+b2)= ( )A.-1或2B.0或2C.2D.1【解析】选 C.由题意可知,a n+1+a n-1=2a n=,解得a n=2(n≥2)(由于数列{a n}每项都是正数),又b n+1b n-1==2b n(n≥2),所以b n=2(n≥2),log2(a2+b2)=log24=2.4.《莱因德纸草书》是世界上最古老的数学著作之一.书中有这样一道题目:把100个面包分给5个人,使每人所得成等差数列,且使较大的三份之和的是较小的两份之和,问最小的一份为( )A. B. C. D.【解析】选 A.设五个人所分得的面包为a-2d,a-d,a,a+d,a+2d(其中d>0),则(a-2d)+(a-d)+a+(a+d)+(a+2d)=5a=100,所以a=20,由(a+a+d+a+2d)=a-2d+a-d,解得d=,所以最小1份为a-2d=20-=.5.(2016·洛阳模拟)在1到104之间所有形如2n和3n(n∈N*)的数,它们各自之和的差的绝对值为( )A.1 631B.6 542C.15 340D.17 424【解析】选B.由2n<104,得n<≈≈13.29,故数列{2n}在1到104之间的项共有13项,它们的和S1==16382;同理数列{3n}在1到104之间的项共有8项,它们的和S2==9840,所以|S1-S2|=6542.【加固训练】1.将正奇数数列各项从小到大依次排成一个三角形数阵:13 57 9 1113 15 17 19……………记M(s,t)表示该表中第s行的第t个数,则表中的奇数2007对应于( )A.M(45,14)B.M(45,24)C.M(46,14)D.M(46,15)【解题提示】先求出2007在原数列中的位置后再求解.【解析】选A.由表知前n行共有1+2+3+…+n=项,当n=44时有990项,又表中的奇数2007是原数列的第1004项,因此2007位于表中第45行的第14个位置.2.(2016·成都模拟)已知数列{a n}满足3a n+1+a n=4(n≥1),且a1=9,其前n项和为S n,则满足不等式|S n-n-6|<的最小整数n的值为( )A.5B.6C.7D.8【解析】选 C.由已知式子变形得3(a n+1-1)=-(a n-1),则{a n-1}是以8为首项,-为公比的等比数列,则|S n-n-6|=|a n-1+a n-1-1+…+a1-1-6|==6×<,化简得3n-1>250,故满足条件的最小整数n的值为7.二、填空题(每小题5分,共15分)6.(2016·茂名模拟)各项都是正数的等比数列的公比q≠1,且a2,a3,a1成等差数列,则的值为.【解析】{a n}的公比为q(q>0且q≠1),由a3=a2+a1,得q2-q-1=0,解得q=,而===.答案:7.(2016·常德模拟)已知数列{a n}的前n项和为S n,对任意n∈N*都有S n=a n-,若1<S k<9(k∈N*),则k的值为.【解析】因为S n=a n-,所以S1=a1-=a1,a1=-1.a n=S n-S n-1(n>1),即a n=-=a n-a n-1,整理得:=-2(n>1),所以{a n}是首项为-1,公比为-2的等比数列,S k==,因为1<S k<9,所以1<<9,即4<(-2)k<28,仅当k=4时成立.答案:48.(2016·新乡模拟)如图,坐标纸上的每个单元格的边长为1,由下往上的六个点:1,2,3,4,5,6的横、纵坐标分别对应数列{a n}(n∈N*)的前12项(如下表所示),按如此规律下去,则a2017+a2018+a2019= .【解析】a1=1,a2=1,a3=-1,a4=2,a5=2,a6=3,a7=-2,a8=4等,这个数列的规律是奇数项为1,-1,2,-2,3,-3,…,偶数项为1,2,3,…,故a2017+a2019=0,a2018=1009.答案:1009(20分钟40分)1.(5分)(2016·福州模拟)已知a,b,c成等比数列,a,m,b和b,n,c分别成两个等差数列,则+等于( )A.4B.3C.2D.1【解析】选C.由题意得b2=ac,2m=a+b,2n=b+c,则+====2.【一题多解】解答本题,还有以下解法:特殊值法:选C.因为a,b,c成等比数列,所以令a=2,b=4,c=8,又a,m,b和b,n,c分别成两个等差数列,则m==3,n==6,因此+=+=2.2.(5分)学校餐厅每天供应500名学生用餐,每星期一有A,B两种菜可供选择.调查表明,凡是在这星期一选A菜的,下星期一会有20%改选B菜;而选B菜的,下星期一会有30%改选A菜,用a n表示第n个星期一选A 菜的人数,如果a1=428,则a4的值为( )A.324B.316C.304D.302【解析】选B.依题意有:a n=a n-1+(500-a n-1)=a n-1+150(n≥2,n∈N*),即a n-300=(a n-1-300)(n≥2,n∈N*),a n=128·+300,因此a4=128·+300=316.【加固训练】根据市场调查结果,预测某种家用商品从年初开始的n个月内累积的需求量S n(单位:万件)近似地满足S n=(21n-n2-5)(n=1,2,…,12).按此预测,在本年度内,需求量超过1.5万件的月份是( )A.5月,6月B.6月,7月C.7月,8月D.8月,9月【解析】选C.设第n个月的需求量为a n,因为从年初开始的n个月内累积的需求量为S n(n=1,2,3,…,12). 所以当n≥2时,a n=S n-S n-1=(21n-n2-5)-[21·(n-1)-(n-1)2-5]=(-n2+15n-9).当n=1时,a1=S1=,适合上式,综上可知,a n=(-n2+15n-9).令a n>1.5,即(-n2+15n-9)>1.5,解得6<n<9.又n的取值为1,2,3,…,12,所以n=7或n=8.3.(5分)(2016·太原模拟)已知函数f(x)=log2x,若数列{a n}的各项使得2,f(a1),f(a2),…,f(a n),2n+4成等差数列,则数列{a n}的前n项和S n= .【解析】设等差数列的公差为d,则由题意,得2n+4=2+(n+1)d,解得d=2,于是log2a1=4,log2a2=6,log2a3=8,…,从而a1=24,a2=26,a3=28,….易知数列{a n}是等比数列,其公比q==4,所以S n==(4n-1).答案:(4n-1)4.(12分)(2016·怀化模拟)已知数列{a n}的前n项和为S n,且满足a1=,a n=-2S n·S n-1(n≥2).求证:++…+≤-.【证明】因为a n=-2S n·S n-1(n≥2),所以S n-S n-1=-2S n·S n-1(n≥2).两边同除以S n·S n-1,得-=2(n≥2),所以数列是以==2为首项,以d=2为公差的等差数列,所以=+(n-1)·d=2+2(n-1)=2n,所以S n=.将S n=代入a n=-2S n·S n-1,得a n=因为=<=(n≥2),=,所以当n≥2时,++…+=++…+<++…+=-;当n=1时,==-.综上,++…+≤-.【加固训练】(2016·唐山模拟)已知公差不为0的等差数列的前n项和为S n,S5=25,且a2,a5,a14成等比数列.(1)求数列的通项公式.(2)设b n=,T n=b1·b2·b3·…·b n,求证:T n≥.【解析】(1)设公差为d(d≠0),因为S5=25,所以5a1+d=25,即a1+2d=5.因为a2,a5,a14成等比数列,所以(a1+d)(a1+13d)=(a1+4d)2,即6a1d-3d2=0.因为d≠0,所以d=2a1,所以a1=1,d=2.则a n=a1+(n-1)d=2n-1.(2)因为b n==,所以T n=b1·b2·b3·…·b n=×××…×.因为>,n∈N*,所以当n≥2时,=××××××…××>××××××…××=.即T n>,n≥2.又当n=1时,T1=≥=成立,综上,当n∈N*时,T n≥成立.5.(13分)(2016·宜昌模拟)我们在下面的表格内填写数值:先将第1行的所有空格填上1;再把一个首项为1,公比为q的数列依次填入第一列的空格内;然后按照“任意一格的数是它上面一格的数与它左边一格的数之和”的规则填写其他空格.(1)设第2行的数依次为B1,B2,…,B n,试用n,q表示B1+B2+…+B n的值.(2)设第3列的数依次为c1,c2,c3,…,c n,求证:对于任意非零实数q,c1+c3>2c2.【解析】(1)B1=q,B2=1+q,B3=1+(1+q)=2+q,…,B n=(n-1)+q,所以B1+B2+…+B n=1+2+…+(n-1)+nq=+nq.(2)c1=1,c2=1+(1+q)=2+q,c3=(2+q)+(1+q+q2)=3+2q+q2,由c1+c3-2c2=1+3+2q+q2-2(2+q)=q2>0,得c1+c3>2c2.。

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课时提升作业(三十四)一、选择题1.等差数列{a n }的公差为3，若a 2,a 4,a 8成等比数列，则a 4=( ) (A)8 (B)10 (C)12 (D)162.等差数列｛a n ｝的公差不为零，首项a 1＝1，a 2是a 1和a 5的等比中项，则数列的前10项之和是( )(A)90 (B)100 (C)145 (D)1903.(2013·济南模拟)某火箭在点火后第一秒钟通过的路程为2 km,以后每秒钟通过的路程增加2 km,在到达离地面240 km 的高度时，火箭与飞船分离，则这一过程需要的时间为( )(A)10秒钟 (B)13秒钟 (C)15秒钟 (D)20秒钟4.(2013·石家庄模拟)《莱因德纸草书》是世界上最古老的数学著作之一.书中有一道这样的题目：把100个面包分给五个人，使每人所得成等差数列，且使较大的三份之和的17是较小的两份之和，问最小一份为( ) (A)53(B)103(C)56 (D)1165.(2013·海淀模拟)已知数列{a n }满足：a 1=1,a n >0,22n 1n a a +-=1(n ∈N *)，那么使a n <5成立的n 的最大值为( )(A)4 (B)5 (C)24 (D)25 6.(2013·合肥模拟)已知数列{a n }为等差数列，公差为d,若1110a a ＜-1，且它的前n项和S n有最大值，则使得S n＜0的n的最小值为( )(A)11 (B)19 (C)20 (D)217.在1到104之间所有形如2n和3n(n∈N*)的数，它们各自之和的差的绝对值为(lg 2≈0.301 0)( )(A)1 631 (B)6 542 (C)15 340 (D)17 4248.(能力挑战题)甲、乙两间工厂的月产值在2012年元月份时相同,甲以后每个月比前一个月增加相同的产值.乙以后每个月比前一个月增加产值的百分比相同.到2012年11月份发现两间工厂的月产值又相同.比较甲、乙两间工厂2012年6月份的月产值大小，则有( )(A)甲的产值小于乙的产值 (B)甲的产值等于乙的产值(C)甲的产值大于乙的产值 (D)不能确定二、填空题9.(2013·温州模拟)设曲线y=x n(1-x)在x=2处的切线与y轴交点的纵坐标为a n,则数列{n a}的前n项和S n等于___________.n110.从盛满2升纯酒精的容器里倒出1升纯酒精，然后填满水，再倒出1升混合溶液后又用水填满，以此继续下去，则至少应倒________次后才能使纯酒精体积与总溶液的体积之比低于10%.11.(2013·山东师大附中模拟)已知S n是等差数列{a n}的前n项和，S10＞0并且S11=0，若S n≤S k对n∈N*恒成立，则正整数k构成的集合为_______.12.(能力挑战题)数列{a n}的前n项和记为S n，a1=t，点(S n,a n+1)在直线y=2x+1上，n∈N*，若数列{a n}是等比数列，则实数t=__________.三、解答题13.等比数列{a n }的前n 项和为S n ，已知S 1,S 3,S 2成等差数列， (1)求{a n }的公比q. (2)若a 1-a 3＝3，求S n .14.(2012·安徽高考)设函数f(x)=x 2+sin x 的所有正的极小值点从小到大排成的数列为{x n }.(1)求数列{x n }的通项公式.(2)设{x n }的前n 项和为S n ，求sin S n .15.(2013·宁波模拟)在等差数列{a n }中，a 1=3，其前n 项和为S n ，等比数列{b n }的各项均为正数，b 1=1，公比为q ，且b 2+S 2=12，q=22S b . (1)求a n 与b n . (2)证明：12n 111123S S S 3≤+++<….答案解析1.【解析】选C.令首项为a ，根据条件有(a+9)2=(a+3)〃(a+21)⇒a=3， a 4=3+3〓3=12.故选C.2.【解析】选B.设公差为d ，则(1+d)2=1〃(1+4d). ∵d ≠0，解得d ＝2，∴S 10＝100.3.【解析】选C.设每一秒钟通过的路程依次为a 1,a 2,a 3,…,a n ，则数列{a n }是首项a 1=2,公差d=2的等差数列，由求和公式有1n(n 1)dna 2402-+=,即2n+n(n-1)=240,解得n=15.4.【解析】选A.设五个人所分得的面包为a-2d,a-d,a,a+d,a+2d(其中d ＞0)，则(a-2d)+(a-d)+a+(a+d)+(a+2d)=5a=100,∴a=20. 由17(a+a+d+a+2d)=a-2d+a-d,得3a+3d=7(2a-3d)， ∴24d=11a,∴d=556， 所以，最小的一份为a-2d=20-110563=. 5.【解析】选C.由a 1=1,a n >0,22n 1n a a +-=1(n ∈N *)可得2n a n =,即n a =,要使a n <5，则n<25,故选C.6.【思路点拨】解答本题首先要搞清条件“1110a a ＜-1”及“S n 有最大值”如何使用，从而列出关于a 1,d 的不等式组，求出1a d的取值范围，进而求出使得S n ＜0的n 的最小值，或者根据等比数列的性质求解.【解析】选C.方法一：由题意知d ＜0,a 10＞0,a 11＜0,a 10+a 11＜0,由111a 9d 0,a 10d 0,2a 19d 0,d 0,+⎧⎪+⎪⎨+⎪⎪⎩＞＜＜＜得1a 1992d --＜＜.∵()2n 11n n 1d dS na d n (a )n 222-=+=+-, 由S n =0得n=0或n=1-12a d. ∵12a 19120d-＜＜, ∴S n ＜0的解集为{*12a n N |n 1d∈-＞}, 故使得S n ＜0的n 的最小值为20.方法二：由题意知d ＜0,a 10＞0,a 11＜0,a 10+a 11＜0,由a 10＞0知S 19＞0,由a 11＜0知S 21＜0, 由a 10+a 11＜0知S 20＜0,故选C. 7.【解析】选B.由2n <104，得44n lg20.301 0<≈≈13.29，故数列{2n }在1到104之间的项共有13项，它们的和1312(21)S 16 38221-==-；同理数列{3n }在1到104之间的项共有8项，它们的和823(31)S 9 84031-==-， ∴|S 1-S 2|=6 542.8.【解析】选C .设甲各个月份的产值构成数列{a n }，乙各个月份的产值构成数列{b n }，则数列{a n }为等差数列，数列{b n }为等比数列，且a 1=b 1,a 11=b 11，故a 6=1116a a b 2+≥===，由于在等差数列{a n }中的公差不等于0，故a 1≠a 11，上面的等号不能成立，故a 6>b 6，即6月份甲的产值大于乙的产值. 9.【解析】∵y ′=nx n-1-(n+1)x n ,∴y ′|x=2=n 〃2n-1-(n+1)〃2n =-n 〃2n-1-2n , ∴切线方程为y+2n =(-n 〃2n-1-2n )(x-2), 令x=0得y=(n+1)〃2n ，即a n =(n+1)〃2n , ∴n n a 2n 1=+,∴S n =2n+1-2. 答案:2n+1-210.【解析】设开始纯酒精体积与总溶液体积之比为1，操作一次后纯酒精体积与总溶液体积之比11a 2=，设操作n 次后，纯酒精体积与总溶液体积之比为a n ,则a n+1=a n 〃12, ∴n 1n n n 1111a a q (),()2210-==∴<，得n ≥4. 答案:4【方法技巧】建模解数列问题对于数列在日常经济生活中的应用问题，首先分析题意，将文字语言转化为数学语言，找出相关量之间的关系，然后构建数学模型，将实际问题抽象成数学问题，明确是等差数列问题、等比数列问题，是求和还是求项，还是其他数学问题，最后通过建立的关系求出相关量.11.【解析】在等差数列{a n }中，由S 10＞0,S 11=0得，()110101105610a a S 0a a 0a a 0,2+=⇒+⇒+＞＞＞ ()11111111611a a S 0a a 2a 0,2+==⇒+==故可知，等差数列{a n }是递减数列且a 6=0, 所以S 5=S 6≥S n ,即k=5或6. 答案:{5，6}12.【思路点拨】得出关于a n+1,S n 的式子，降低一个角标再得一个关于a n ,S n-1的式子，两个式子相减后得出a n+1,a n 的关系，可得数列{a n }中，a 2,a 3,a 4,…为等比数列，只要21a a 等于上面数列的公比即可. 【解析】由题意得a n+1=2S n +1, a n =2S n-1+1(n ≥2),两式相减得a n+1-a n =2a n ，即a n+1=3a n (n ≥2), 所以当n ≥2时，{a n }是等比数列， 要使n ≥1时，{a n }是等比数列，则只需21a 2t 1a t+==3，从而t=1. 答案:113.【解析】(1)依题意有a 1+(a 1+a 1q)=2(a 1+a 1q+a 1q 2),由于a 1≠0，故2q 2+q=0. 又q ≠0，从而q=-12. (2)由已知可得a 1-a 1(-12)2=3, 故a 1=4,从而n n n 14[1()]812S [1()]1321()2--==----. 14.【思路点拨】(1)根据导数，x n 的左侧导函数小于0，x n 的右侧导函数大于0，求出极小值点.(2)由(1)求出{x n }的前n 项和为S n ，再代入sin S n 求解. 【解析】(1)f(x)=x 2+sin x,令f'(x)=12+cos x=0,得x=2k π〒23π(k ∈Z), f'(x)>0⇒2k π-23π<x<2k π+23π(k ∈Z), f'(x)<0⇒2k π+23π<x<2k π+43π(k ∈Z),当x=2k π-23π(k ∈Z)时，f(x)取极小值,x n =2n π-23π(n ∈N *).(2)由(1)得：x n =2n π-23π,S n =x 1+x 2+x 3+…+x n =2π(1+2+3+…+n)-2n 3π=n(n+1)π-2n 3π.当n=3k(k ∈N *)时，sin S n =sin(-2k π)=0, 当n=3k-1(k ∈N *)时，sin S n =sin 23π=2, 当n=3k-2(k ∈N *)时，sin S n=sin43π=.所以**n *0,n 3k,k N ,sin S n 3k 1,k N ,n 3k 2,k N .⎧⎪=∈==-∈⎪⎪=-∈⎪⎩15.【解析】(1)设{a n }的公差为d ，因为2222b S 12,S q ,b +=⎧⎪⎨=⎪⎩所以q 6d 12,6d q ,q ++=⎧⎪+⎨=⎪⎩ 解得q=3或q=-4(舍)，d=3. 故a n =3+3(n-1)=3n ，b n =3n-1. (2)因为n n(33n)S 2+=， 所以()n 12211()S n 33n 3n n 1==-++. 故12n111S S S +++… =2111111121[(1)()()()](1)322334n n 13n 1-+-+-+⋯+-=-++.因为n ≥1，所以110n 12<≤+，于是11112n 1≤-<+， 所以1212(1)33n 13≤-<+. 即12n 111123S S S 3≤++⋯+<. 关闭Word 文档返回原板块。

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课时提升作业（八）等差数列（25分钟60分）一、选择题（每小题5分，共25分）l.x+1与yT 的等差中项为10,则x+y 等于（）【解析】选C.因为x+1与y-1的等差中项为10, 所以（x+l ） + （y-l ）=2X10, 所以 x+y 二20.2. （2015 -长沙高一检测）已知等差数列｛爲｝满足创二0, a 6+a 8=-10,则 &2016二（ ） A. 2014B. 2015C.-2014D. -2015【解析】选C.设等差数列｛&］的公差为d,则由已知条件可得 即裁j 解得&==-£所以数列心的通项公式为 Qn 二-n+2 ,故 32016~—2014. 【补偿训练】（2015 •吉安高二检测）在等差数列｛a 」中，若a 2=-5,&6二81+6,则 810 等于（ ）【解析】选A.由题意，得25分钟基础练＞A. 0B. 10C. 20D.不确定 A. 19B. 18C.-19D.-18严+ d = —5, 严=7 & + 5d = a 】+ 3d + 6, Id = 3,所以 a n =3n-11,所以 a w =19.3. （2015 -大连高二检测）在数列｛务｝中，屮2, 2亦一2箱1,则咖的 值为（）【解析】选A •因为2a n+-2a n =l, 所以 a n+i-a n =^,所以数列｛aj 是首项为2,公差为g 的等差数列， 所以 a ⑹二2+（101T ） X ；二52.4. （2015 •东营高二检测）首项为-24的等羌数列，从第10项起开始【误区警示】解决本题时容易忽视前9项是小于等于零的条件而选A.5•在等差数列-5, -3*, -2, W ，…中，每相邻两项之间插入一个数,使之组成一个新的等差数列，则新数列的通项公式为（）A. d>|3B. d<3 C ・？Wd 〈338D.E 〈dW33【解析】选 D ・设公差为d, a n =-24+(n-l)d.由题知I®9可得U1Q > E为正数，则公差d 的取值范围为（）得|〈dW3・A. 52B. 51C. 50D. 49—24 + 8d 吒(X —24 + 9d > 0,B. a n=-5-| (n~l)【解题指南】解答本题的关键是确定新等差数列的公差，实际上新数列的公差为原数列公差的一半.【解析】选A.首项为-5,公差为主上」，z 4“宀_匚、(八3_3 23所以a n——5+ (n—1) • —n—•4 4 4二、填空题(每小题5分，共15分)6. (2015 -五指山高二检测)已知等差数列&｝的前三项为a-1, a+1,2a+3,则此数列的通项公式为___________ ・【解析】因为a-1, a+1, 2a+3成等差数列，所以2 (a+1) -a_1 +2a+3,解得a-0.等差数列｛a」的前三项为-1,1, 3,其首项为T,公差为2,所以a n——1+ (n-1) X 2—2n—3.答案:a…=2n-37•若xHy,两个数列：x, a,, a2, a3, y 和x, b】，b2, b3, b4, y 都是等差数列，则—的值为________________________ ・【解析】设两个等差数列的公差分别为d], d2,【补偿训练】在-1和8之间插入两个数3, b(a<b),使这四个数成等即求学，由已知得答案订差数歹!b贝I」a二____ , b二________ .【解析】d二字空二3,所以a二-1+3=2, b二2+3二5・答案：2 58•在数列｛&」中，ai=3,对于任意大于1的正整数n,点(《乔、® 在直线x-y- V3=0上，则a,= __________________ ・【解析】由题意，得V^-vaT7=V3(n^2),又aF3,所以数列｛、：'瓦'｝是以\总为首项，w逗为公差的等差数列，所以、瓦"二\'3+ (n-1) X <3= v^n,所以a“二3nl答案：3n2三、解答题(每小题10分，共20分)9.在等差数列｛a」中,ai+a5=8, a4=7.(1)求数列的第10项.⑵问112是数列｛&｝的第几项?⑶数列｛a n｝从第几项开始大于30?⑷在80到110之间有多少项？【解析】设｛a」公差为d,则｛；：：；壮笄'解得占二厂(1)a10=a1+9d=-2+27=25.(2)a n=_2+ (n_1) X 3-3n_5,由112=3n-5,解得n二39.所以门2是数列｛aj的第39项.2(3)令3n-5>30 解得n>11-,所以从第12项开始大于30.(4)由80<3n-5<110,解得1 128-<n<38-,3 3’所以n的取值为29, 30,…，38,共10项.10.一位同学喜欢观察小动物的活动规律，他观察到随着气温的升高, 一种昆虫在相等的时间内发出的碉啾声次数也在逐渐增加•下表是他记录的数据，34上方及40下方的数据变得模糊不清了•但是该同学记得气温每升高rc他观察一次，而且观察到的数据成等差数列•请你为他补好这两个数据.【解析】设昆虫阴啾声次数组成等差数列｛a」，则3i~4, 35~20,温度为34°C时，勺二a〔+6d・又因为d仝厂屯-兰二4,所以a7=4+6 X 4二28.4 4若an二40,则4+(n-1) X4=40.所以n=10,所以温度为37°C.【补偿训练】某公司经销一种数码产品，第1年获利200万元，从第2年起由于市场竞争等方面的原因，利润每年比上一年减少20万元, 按照这一规律如果公司不开发新产品，也不调整经营策略，从哪一年起，该公司经销这一产品将亏损？【解析】由题意可知，设第1年获利为a1?第n年获利为a n,则a n-a n_F-20 (n^2,n£N*),每年获利构成等差数列｛a」，且首项3^200, 公差d二-20,所以a n-ai+ (n~1) d =200+ (n-1) X (-20)二-20n+220.若a n<0,则该公司经销这一产品将亏损，由a=-20n+220<0,解得n>11, 即从第12年起，该公司经销这一产品将亏损.⑳分钟提升练'(20分钟40分)一、选择题(每小题5分，共10分)1. (2015 •常德高二检测)已知等差数列｛a n｝的公差dHO,且a：产2a,则◎的值为()A.-B.-6 5【解题指南】由题意可得d和內的关系，可得通项公式，代入要求的式子化简.【解析】选C.因为等差数列｛a」的公差dfO,且a3=2ai,所以a3=ai+2d=2ai,所以aF2d,所以a n=2d+ (n-1)d= (n+1)d,所以比+电_加+理屯+鮎3d4?d 42. (2015 -鹰潭高二检测)如图，按英文字母表A, B, C, D, E, F, G,II,…的顺序有规律排列而成的鱼状图案中，字母“0”出现的个数为VCCC9A. 27B. 29C. 31D. 33【解析】选B.由题意可得字母A有1个，B有3个，C有5个，D有7个…，它们构成以1为首项，2为公差的等差数列，所以通项公式为a n-1+2 (n_l) =2n_l,因为字母“0”在第15个，所以字母“0”出现的个数315=2X15-1=29.二、填空题(每小题5分，共10分)3•数列｛a」是等差数列，&1与出的等差中项为1, a?与％的等慕中项为2,则公差d二 __________ .【解析】由题意得81+^2二2, a2+a3=4,所以(a2+a3)-(a〔+a2)二4-2二2,所以a3-ai=2,即2d=2,所以d=1・答案：1【补偿训练】若m和2n的等差中项为4, 2m和n的等差中项为5, 则m与n的等差中项是 _______________ ・【解析】因为m和2n的等差中项为4,所以m+2n=8.因为2m和n的等差中项为5,两式相加，得3m+3n=18,即m+n二6,故m 与n 的等差中项为巴竺二E 二3.答案：34. (2015 •遵义高一检测)已知在数列｛a n ｝中，ai=-l, a n+i • a n =a n+i-a n , 则数列通项a n =【解析】由题意可知a n *0, nWN ；所以由 a n+i • a n =3n +i —3n,1 1 两边同除以a n+1 - a n ,整理得 =-1,a n^i a n 所以数列｛十］是首项为T,公差为T的等差数列，11所以一+ (n-1) X (T)二一n,所以 a n =—• a n n 答案：-丄 n【延伸探究】将木题条件改为屮1, a 24, 结果又如何?2 a 】出a n a 】*2 i 1 【解析】由已知 --- —h --- 可得 1_ 1! _________ 1 an 十』&n+i i i 公差为 ---- =2-1=1的等差数列， 1 1所以一=1 + (n-1) X 1=n,故 a n —. a n n【拓展延伸】构造辅助数列巧求数列通项公式 观察递推公式的特征，构造恰当的辅助数列使之转化为等差数列问题. 常用方法有: 平方法、开平方法、倒数法等•例如， 数列｛a 」中，ai-1, a n +i — ,求 3n .a (t+2此题可取倒数，构造辅助数列｛彩｝来求解1是首项为一=1 ,315. 已知数列｛a 」满足：ai=10, a 2=5, a n -a n .2=2 (neN*).求数列｛a 」的通 项公式.【解析】因为 3i —10, 32—5, a n —a n +2~2 (n £ N ),所以数列｛a 」的奇数项、偶数项均是以-2为公差的等差数列. 当 n 为奇数时，a“二ai+(—T* — 1) X (-2) =11-n, 当 n 为偶数时，a=a 2+(^- 1)X (-2)=7-n,■ 一 4 F ・为奇数* 7 - n. n 为偶数. 6. (2015 •临沂高二检测)已知数列｛a,J 中， Z 数列｛捕满足亦話(心)•(1)求证:数列｛bn ｝是等差数列. 11又 bF^-=-ai-1 所以数列｛bj 是以三为首项，以1为公差的等差数列.7 1⑵由⑴知，b n =n--,则a=1+—2如2 =1+韵，设函数f (x )h+乔7， 所以an 二3 1&i 二二，為二| (n 2 2, n ⑵求数列｛缶｝中的最大值和最小值，并说明理由.【解析】(1)因为务=2 ------ (n^2, nEN*), b n =— a n-i 1 -|i所以当 n $2 时，b n -b n -i= ----- --------1 _ a n.-i 1_d—+ S)内为减函数.易知f (x)在区间当n二4时，令取得最大值3.【补偿训练】数列｛a」满足a n+1=3a…+n(nGN*)，问是否存在适当的使其是等差数列？【解题指南】假设存在，利用等差数列的定义求解确定.【解析】假设存在这样的4满足题目条件.a n+2=3a n+i+n+1 (n G N*).所以a n+2_an+i=2a n+i+n+1,由已知a n+i=3a n+n (n G N*)可得3n+i—3n—2a n+n,所以2a n+i+n+1-2a n+n,所以a n+1-a=4，满足等差数列的定义，故假设是正确的•即存在适当—的內的值使数列｛a」为公差为冷的等差数列.由已知条件a”i二3an+n,令n二1,所以a2=3ai+1,即ai~=3ai+1,解得。

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课时提升作业(五)Ⅰ. 单项填空1. As with many women of her age, she struggled to find a between her career and her family.A. balanceB. connectionC. answerD. compromise2. (2013·天津模拟)As far as we are concerned, education is about learning and the more you learn, .A. the more for life are you equippedB. the more equipped for life you areC. the more life you are equipped forD. the more life for you are equipped3. Experts warn that global warming will cause extreme climate changes including more floods, heat waves and droughts.A. usualB. ordinaryC. frequentD. common4. The play next month aims mainly the local culture.A. produced; to reflectB. being produced; at reflectingC. to be produced; to reflectD. to be produced; reflecting5. (2013·绍兴模拟)You’d sound a lot more polite if you make a request a question.A. in search ofB. in the form ofC. in need ofD. in the direction of6. (2013·重庆模拟)—Bill, could I use your phone for a while?—. Go ahead, please.A. By all meansB. Couldn’t be betterC. With pleasureD. No wonder7. —Liu Xiang is on the road to recovery; he needs improvement step by step.—I hope injuries will from him.A. send awayB. keep awayC. put awayD. take away8. (2013·贵阳模拟)was careless him to leave his umbrella in the store.A. It; forB. It; ofC. That; forD. That; of9. The pleasant weather our pleasure. We had a wonderful time during our holiday.A. added up toB. added inC. added upD. added to10. (2013·桂林模拟)It is reported that the government has the outbreak of violence by sending a large group of soldiers to police the area.A. gone forB. reacted toC. appealed toD. made for11. I’m used to in the house which to keep books and magazines.A. living; usesB. live; was usingC. live; usedD. living; used12. It’s said that within five years cars will be a lot in our country.A. cheaperB. cheapestC. cheapD. more cheap13. , I would like to thank everyone who helped to make this event possible.A. In conclusionB. In funC. In wordD. In case14. (2013·大庆模拟)Alice was supposed at home, but her mother found her surfing the Internet.A. studyingB. to be studyingC. to have studiedD. studied15. A computer is the most important piece of you will buy.A. furnitureB. toolC. equipmentD. technologyⅡ. 完形填空(2013·南昌模拟) It was at a concert that I happened to find David. He was holding his head at a strange1as if he were staring down at something. Then the truth struck me. He was blind. The last thing he remembers2was his daughter being born. Then the world went3.Bad luck is no stranger to this 44-year-old man. His mother died of cancer, andhis weak father had to411-year-old David to the care of the state.Things seemed always to go from bad to5. Two years ago, his beloved guide dog pulled him out of the6of a truck. David was not hurt. The dog7.But David does not feel sorry for himself. “These are just little obstacles(障碍)you have to8in your life, ”he said.He has to make a daily two-hour trip to his working place—the X-ray department of an9room.It was a hard job to10. Before he got it, David was determined to escape the workshop run by the Lighthouse, an organization11to help blind people. He wanted a job of developing X-ray film, something12, not just he, must do in the dark, including people with13eyesight. The Lighthouse called many hospitals, with no result, 14they offered to pay his first three months’ salary.David works alone in a dark room that15of chemicals. He cannot wear gloves. 16, he could not feel. Since this is an emergency room, lives can be put 17or out of danger. His directress says she trusts him 100 percent.He makes $20, 000 a year. But his motivation(动机)goes18money. “By working, I can realize my own19. That’s the most important!” he said.What a shining example for us to20!(283W)1. A. ankle B. direction C. angle D. altitude2. A. doing B. feeling C. seeing D. hearing3. A. blank B. colorful C. empty D. silent4. A. bring up B. bring inC. give upD. give in5. A. worse B. good C. better D. mad6. A. bottom B. path C. wheel D. back7. A. killed B. sacrificedC. diedD. injured8. A. seek B. overcomeC. surroundD. search9. A. upstairs B. emergencyC. informationD. underground10. A. deal with B. get throughC. take upD. come by11. A. intended B. affordedC. offeredD. devoted12. A. everyone B. nobodyC. someoneD. none13. A. common B. poorC. normalD. actual14. A. now that B. as ifC. in caseD. even though15. A. consists B. smellsC. allowsD. fills16. A. However B. UnfortunatelyC. OtherwiseD. Besides17. A. at risk B. in briefC. up withD. in charge18. A. into B. beyond C. onto D. below19. A. blindness B. incomeC. valueD. experiences20. A. set B. take C. hold D. followⅢ. 阅读理解(Nov. 16, 2012)—Children who learn how to swim at ayoung age are reaching many developmental milestones earlierthan the norm.Researchers from the Griffith Institute for EducationalResearch surveyed parents of 7, 000 under-fives from Australia,New Zealand and the US over three years.A further 180 children aged 3, 4 and 5 have been involved in intensive testing, making it the world’s most comprehensive study into early-years swimming.Lead researcher Professor Robyn Jorgensen says the study shows young children who participate in early-years swimming achieve a wide range of skills earlier than the normal population.“Many of these skills are those that help young children into the transition(过渡, 转变)into formal learning surroundings such as pre-school or school. ”The research also found significant differences between the swimming groups and non-swimmers regardless of socio-economic background.While the two higher socio-economic groups performed better than the lowertwo in testing, the four SES(美国工程科学协会)groups all performed better than the normal population.The researchers also found there were no gender differences between the research groups and the normal population.As well as achieving physical milestones faster, children also scored significantly better in visual-motor skills such as cutting paper, colouring in and drawing lines and shapes, and many mathematically-related tasks. Their oral expression was also better as well as in the general areas of literacy and numeracy.The study is a joint project between Griffith University, Kids Alive Swim Program and Swim Australia. (281W)1. The purpose of the passage is.A. to advertise for swimmingB. to call on children to learn how to swimC. to illustrate a study related to swimmingD. to survey those children who swam at an early age2. Which of the following is the result of the study?A. More than 7, 000 children participated in the testing.B. The younger children learn to swim, the wiser they are.C. Boys are cleverer than girls.D. The scores of the two higher socio-economic groups are higher than the lower two in testing.3. How many s kills that children achieved in their early years’swimming are mentioned in the passage?A. 3.B. 4.C. 5.D. 6.4. Where can you probably find the text?A. In a popular magazine.B. In a Science Daily.C. In a physics textbook.D. In an official report.【语篇随练】请运用上下文暗示法, 按照以下提示题号找出完形填空中的对应词答案解析Ⅰ.1.【解析】选A。

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课时提升作业(五)同角三角函数的基本关系(25分钟 60分)一、选择题(每小题5分，共25分)1.sin α=√55，则sin 2α-cos 2α的值为( )A.-15B.-35C.15D.35【解析】选B.由于sin α=√55，所以cos 2α=1-sin 2α=45，则原式=15-45=-35.【延长探究】本题条件下，求sin 4α-cos 4α的值. 【解析】由sin 4α-cos 4α=(sin 2α+cos 2α)(sin 2α-cos 2α)=sin 2α-cos 2α =-35.2.(2021·福建高考)若sin α=-513，且α为第四象限角，则tan α的值等于( )A.125B.-125C.512D.-512【解题指南】利用同角三角函数关系，“知一求二”.【解析】选D.由sin α=-513，且α为第四象限角可知cos α=1213，故tan α=sinαcosα=-512.3.(2021·葫芦岛高一检测)已知α是其次象限角，cos α=-13，则3sin α+tan α=( )A.-√2B.√2C.-1D.0 【解析】选D.由于cos α=-13，α是其次象限角，所以sin α=√1−cos 2α=√1−(−13)2=2√23. 所以tan α=sinαcosα=2√23−13=-2√2.所以3sin α+tan α=3×2√23-2√2=0. 4.(2021·重庆高一检测)已知角θ为第四象限角，且tan θ=-34，则sin θ- cos θ=( )A.15B.75C.-15D.-75【解析】选D.由已知得{sinθcosθ=−34,sin 2θ+cos 2θ=1,所以(−34cosθ)2+cos 2θ=1，cos 2θ=1625，又角θ为第四象限角，所以cos θ=45.所以sin θ=-34cos θ=-34×45=-35. 所以sin θ-cos θ=-35-45=-75.5.已知sin α-cos α=-√52，则tan α+1tanα的值为( )A.-4B.4C.-8D.8【解析】选C.tan α+1tanα=sinαcosα+cosαsinα=1sinαcosα.由于sin αcos α=1−(sinα−cosα)22=-18，所以tan α+1tanα=-8.二、填空题(每小题5分，共15分)6.(2021·北京高一检测)已知α是其次象限的角，且sin α=513，则cos α=________.【解析】由于α是其次象限的角，且sin α=513，所以cos α=-√1−sin 2α=-√1−(513)2=-1213.答案：-12137.若sin θ=k+1k−3，cos θ=k−1k−3，且θ的终边不落在坐标轴上，则tan θ的值为________.【解析】由于sin 2θ+cos 2θ=(k+1k−3)2+(k−1k−3)2=1，所以k 2+6k-7=0，所以k 1=1或k 2=-7.当k=1时，cos θ不符合，舍去. 当k=-7时，sin θ=35，cos θ=45，tan θ=34.答案：348.已知sinx=3cosx ，则sinxcosx 的值是________. 【解析】将sinx=3cosx 代入sin 2x+cos 2x=1中得9cos 2x+cos 2x=1，即cos 2x=110， 所以sin 2x=1-cos 2x=910， 由于sinx 与cosx 同号，所以sinxcosx>0， 则sinxcosx=√sin 2xcos 2x =310.答案：310三、解答题(每小题10分，共20分) 9.(2021·武汉高一检测)已知tan 2α1+2tanα=13，α∈(π2,π). (1)求tan α的值. (2)求sinα+2cosα5cosα−sinα的值.【解析】(1)由tan 2α1+2tanα=13，得3tan 2α-2tan α-1=0，即(3tan α+1)(tan α-1)=0，解得tan α=-13或tan α=1.由于α∈(π2,π)，所以tan α<0，所以tan α=-13.(2)由(1)，得tan α=-13，所以sinα+2cosα5cosα−sinα=tanα+25−tanα=−13+25−(−13)=516.【延长探究】本例条件下，计算sin 2α+sin αcos α的值.【解析】sin 2α+sin αcos α=sin 2α+sinαcosαsin 2α+cos 2α=tan 2α+tanαtan 2α+1=(−13)2+(−13)(−13)2+1=-15.10.求证：3-2cos 2α=3tan 2α+1tan 2α+1.【证明】右边=3(tan 2α+1)−2tan 2α+1=3-2tan 2α+1=3-2sin 2αcos 2α+1=3-2cos 2αsin 2α+cos 2α=3-2cos 2α=左边，所以原式得证. 【一题多解】左边=3(sin 2α+cos 2α)−2cos 2αsin 2α+cos 2α=3sin 2α+cos 2αsin 2α+cos 2α=3tan 2α+1tan 2α+1=右边，所以原式得证.(20分钟 40分)一、选择题(每小题5分，共10分)1.化简sin 2α+cos 4α+sin 2αcos 2α的结果是( ) A.14B.12C.1D.32【解析】选C.原式=sin 2α+cos 2α(cos 2α+sin 2α)=sin 2α+cos 2α=1.【补偿训练】若sin α+sin 2α=1，则cos 2α+cos 4α等于________.【解析】由于sin α+sin 2α=1，sin 2α+cos 2α=1，所以sin α=cos 2α，所以cos 2α+cos 4α=sin α+sin 2α=1. 答案：12.(2021·宣城高一检测)已知sin θ=2cos θ，则sin 2θ+sin θcos θ-2cos 2θ等于( )A.-43B.54C.-34D.45【解题指南】关于sin θ，cos θ的齐次式，可用1的代换、化弦为切求值. 【解析】选D.由于sin θ=2cos θ，所以tan θ=sinθcosθ=2， sin 2θ+sin θcos θ-2cos 2θ =sin 2θ+sinθcosθ−2cos 2θsin 2θ+cos 2θ=tan 2θ+tanθ−2tan 2θ+1=22+2−222+1=45.二、填空题(每小题5分，共10分)3.(2021·龙岩高一检测)化简：α为其次象限角，则cosα√1+tan 2α+√1+sinα1−sinα-√1−sinα1+sinα=__________.【解析】原式=cosα√1+2cos 2α+√(1+sinα)21−sin 2α-√(1−sinα)21−sin 2α=cosα·√1cos 2α+|1+sinαcosα|-|1−sinαcosα|.又由于α为其次象限角，所以cos α<0，1+sin α>0，1-sin α>0， 所以原式=1cosα·1−cosα-1+sinαcosα-(−1−sinαcosα)=-1-1+sinαcosα+1−sinαcosα=-1+−2sinαcosα=-1-2tan α.答案：-1-2tan α 【补偿训练】√1−2sin70°cos70°sin70°−√1−sin 270°=________.【解析】原式=√sin 270°+cos 270°−2sin70°cos70°sin70°−√cos 270°=√(sin70°−cos70°)2sin70°−|cos70°|=|sin70°−cos70°|sin70°−|cos70°|由于sin 70°>cos 70°>0， 所以原式=sin70°−cos70°sin70°−cos70°=1.答案：14.已知关于x 的方程4x 2-2(m+1)x+m=0的两个根恰好是一个直角三角形的一个锐角的正、余弦，则实数m 的值为________. 【解析】设直角三角形中的该锐角为β， 由于方程4x 2-2(m+1)x+m=0中， Δ=4(m+1)2-4·4m=4(m-1)2≥0， 所以当m ∈R 时，方程恒有两实根. 又由于sin β+cos β=m+12，sin βcos β=m4，所以由以上两式及sin 2β+cos 2β=1， 得1+2·m4=(m+12)2，解得m=±√3.当m=√3时，sin β+cos β=√3+12>0，sin β·cos β=√34>0，满足题意， 当m=-√3时，sin β+cos β=1−√32<0，这与β是锐角冲突，舍去. 综上，m=√3. 答案：√3三、解答题(每小题10分，共20分)5.(2021·盐城高一检测)已知sin α+cos α=12(0<α<π)，(1)求sin αcos α.(2)求sin α-cos α.【解析】(1)平方得1+2sin αcos α=14，所以sin αcos α=-38.(2)由(1)式知sin αcos α<0，0<α<π，所以π2<α<π，所以sin α-cos α>0，由于(sin α-cos α)2=1-2sin αcos α=74，所以sin α-cos α=√72.【补偿训练】在△ABC 中，sinA+cosA=15，求(1)sinA ·cosA. (2)tanA. 【解析】(1)由于sinA+cosA=15，所以(sinA+cosA)2=125，即1+2sinAcosA=125，所以sinAcosA=-1225.(2)由于sinA+cosA=15，①A ∈(0，π)，所以A ∈(π2,π)，所以sinA-cosA>0，又由于(sinA-cosA)2=1-2sinAcosA =1-2×(−1225)=4925，所以sinA-cosA=75②联立①②解得，sinA=45，cosA=-35，所以tanA=sinAcosA=45−35=-43.6.已知sin θ=asin φ，tan θ=btan φ，其中θ为锐角，求证：cos θ=√a 2−1b 2−1.【证明】由sin θ=asin φ，tan θ=btan φ，得sinθtanθ=asinφbtanφ，即acos φ=bcos θ，而asin φ=sin θ，得a 2=b 2cos 2θ+sin 2θ，即a 2=b 2cos 2θ+1-cos 2θ， 得cos 2θ=a 2−1b 2−1，而θ为锐角，所以cos θ=√a 2−1b 2−1.关闭Word 文档返回原板块。

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课时提升作业(五)必修1 Unit 5Ⅰ. 单项填空1. (2013·西安模拟)Because of the intense competition, the company has to raise handset subsidies(电话津贴)to keep customers as a.A. rewardB. prizeC. awardD. result2. During enjoying the Mid-autumn Festival, celebrating the harvest is an important reason, but of importance is the whole family’s get-together.A. fairB. reasonableC. equalD. proper3. (2013·南宁模拟)My daughter, all kinds of meat, would not like to eat any vegetable, which worries me a lot.A. devoted toB. addicted toC. accustomed toD. exposed to4. (2013·大同模拟)Mike often attempts to escape whenever he breaks traffic regulations.A. having been finedB. to have been finedC. to be finedD. being fined5. (2013·宁波模拟)In time we reached a stage we had more young readers than old ones.A. whereB. howC. whoD. which6. Shortly after two bombs were exploded on Thursday in the Pakistani city, the mayor went to the spot to the matter himself.A. turn toB. see toC. set toD. refer to7. (2013·南阳模拟)For all these years I have been working for others. I’m hoping I’ll ______my own business someday.A. turn upB. fix upC. set upD. make up8. (2013·新余模拟)It’s important to learn to take a positive attitude life when you are trouble.A. to; underB. of; inC. to; inD. with; in9. you talk to someone or write a message, you show your skills to others.A. At timesB. Some timeC. Every timeD. The time10. (2013·南昌模拟)The governor was by the press for failing to keep a campaign promise.A. destroyedB. beatenC. foughtD. attacked11. (2013·绵阳模拟)you eat the correct foods be able to keep fit and stay healthy.A. Only if; will youB. Only if; you willC. Unless; will youD. Unless; you will12. Edible oil(食用油)safety is a subject we have argued for a long time.A. of whichB. with whichC. about whichD. into which13. (2013·衡阳模拟)After the government forces were defeated, the insurgents(反叛者).A. come into beingB. come into useC. come into effectD. come into power14. Can you tell me when he was and what he is doing at present?A. out of orderB. out of workC. out of questionD. out of season15. (2013·南阳模拟)I paid a visit to him, the great scientist was doing an experiment in the lab.A. For the first timeB. The first timeC. By the timeD. It was the first timeⅡ. 完形填空Everyone has had an event that has changed his or her life. Mine was the1 of my little brother, Matthew. My whole life2with his birth, but sometimes change can be a(n)3thing, even when it’s not expected.When my4first told me that they were having a baby and I would be an older brother, I was full of emotions. I was5and sad at the same time. My whole life was going to change and I wasn’t sure6I was ready. Back then, I did not know what a positive impact this7would have on my life.When Matthew was born I could not stop8. There, lying in bed, was thislittle “creature” . Not until I picked him up did I9the weight of his life on my shoulders, 10the weight of the effect I would have on him. I11felt I had more responsibility, and I was ready to do anything for my little brother.My whole family took12of him, even the cat, and life was somewhat peaceful during those first few weeks. When my13went out for the first time in months, I fed him14. It was just as satisfying for me as for Matthew. I was there 15he rolled over for the first time, and I was there when he16his first steps. I saw his first tooth come in, and I will soon see his first tooth17. I helped feed him his first solid food and I18him say his first word “momma” . He called me “Donadan” 19Jonathan. At least he was20an effort to talk to me. I am too pleased to see him growing up happily.1. A. death B. birth C. growth D. development2. A. changed B. ruined C. passed D. improved3. A. ordinary B. special C. bad D. good4. A. friends B. neighboursC. relativesD. parents5. A. worried B. disappointedC. happyD. friendly6. A. if B. that C. what D. why7. A. affair B. matterC. eventD. business8. A. shaking B. cryingC. smilingD. playing9. A. have B. feelC. findD. discover10. A. or B. so C. thus D. and11. A. happily B. luckilyC. immediatelyD. surprisingly12. A. place B. careC. chargeD. advantage13. A. mother B. fatherC. sisterD. brother14. A. cakes B. bread C. vegetables D. milk15. A. before B. after C. when D. while16. A. walked B. watched C. forced D. took17. A. come out B. turn outC. fall outD. pick out18. A. witnessed B. heardC. noticedD. taught19. A. because of B. instead ofC. in place ofD. in spite of20. A. making B. determiningC. tryingD. givingⅢ. 阅读理解(2013·阜新模拟) The advice offered from any other 82-year-olds might have made young peopleyawn and roll their eyes. But when former South Africanpresident Nelson Mandela advised two dozen local youthleaders in Los Angeles to take education seriously, hisaudience was listening.The famed old man said to the young people that if theyexpected to improve the lives of others in the future, they must work at improving their own lives now. “Education is one of the most important weapons you have, ” Mandela advised. “It will place you in a far better position to serve yourself and your community. ”“The point is, he was young once and rebellious once and he kept his dream alive, just as you each have dreams, ” explained South Africa’s ambassador to t he United States, Sheila Sisulu, as she introduced Mandela to the young crowd.Asked for specific advice about changing society by 21-year-old Ahmed Younis, Mandela suggested that somehow helping arouse more American interest in foreign affairs might be a start.“There is an impression that Americans, in general, have not followed international developments properly, ” Mandela said. “I’m not making that statement myself, but there are serious political analysts who say Americans are not well informed as to what has happened in the world. ”22-year-old Omari Trice said Mandela left him full of enthusiasm. “He’s a person who set the tone for an entire nation, ” said Trice. “You come away feeling you need to be a superman in order to get things done. ” (230W)1. From the passage, we can conclude that American youth.A. are willing to accept the advice from world-famous leadersB. usually think that advice from old people is not worth consideringC. have a good understanding of the oldD. have no intention to improve the lives of others2. Nelson Mandela makes the point in his speech that American youth leaders should.A. improve their own livesB. go to college for better educationC. put more importance on educationD. become interested in foreign affairs3. What Sheila Sisulu said suggested that.A. Nelson Mandela never lost hope in his lifeB. she was especially troublesome when youngC. Nelson Mandela was quite different from American youth when youngD. American youth should be no more rebellious4. From what Trice said, we can know that.A. he wa s greatly impressed and encouraged by Mandela’s speechB. he thought little of Mandela’s speechC. he must be a superman in order to change societyD. he’ll be more interested in international development【语篇随练】1. 将阅读理解中的句子翻译成汉语(1)The famed old man said to the young people that if they expected to improve thelives of others in the future, they must work at improving their own lives now.【译】______________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (2)It will place you in a far better position to serve yourself and your community. 【译】______________________________________________________________ ____________________________________________________________________ 2. 用30个词左右概括阅读理解文章大意____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________答案解析Ⅰ.1.【解析】选A。

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单元评估检测（五）第五章(120分钟150分)一、选择题(本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.已知数列{a n}为等差数列,若a2=3,a1+a6=12,则a7+a8+a9= ( )(A)27 (B)36 (C)45 (D)632.(2013·开封模拟)已知等差数列{a n}的前n项和为S n,且S3=6,则5a1+a7的值为( )(A)12 (B)10 (C)24 (D)63.(2013·南阳模拟)已知数列{a n}是等比数列,其前n项和为S n,若a4=2a3,S4=1,则S8= ( )(A)17 (B)16 (C)15 (D)2564.(2013·吉安模拟)等比数列{a n}的公比q>1,+=3,a1a4=,则a3+a4+a5+a6+a7+a8=( )(A)64 (B)31 (C)32 (D)635.已知数列{a n}满足a1=1,a n+1·a n=2n(n∈N+),则a10= ( )(A)64 (B)32 (C)16 (D)86.已知函数f(x)满足f(x+1)=+f(x),x∈R,且f(1)=,则数列{f(n)}(n∈N+)的前20项的和为( )(A)305 (B)315 (C)325 (D)3357.(2013·黄冈模拟)等差数列{a n}的前n项和为S n,若a3+a9+a15+a17=0,则S21的值是( )(A)1 (B)-1 (C)0 (D)不能确定8.在等差数列{a n}中,a1=-2012,其前n项和为S n.若-=2,则S2012的值等于( ) (A)-2011 (B)-2012(C)-2010 (D)-20139.(2013·宜春模拟)设数列{a n}的前n项和为S n,a1=1,a n=+2(n-1)(n∈N+),若S1+++…+-(n-1)2=2013,则n的值为( )(A)1007 (B)1006 (C)2012 (D)201310.(2013·南昌模拟)已知数列{a n}是各项均为正数且公比不等于1的等比数列.对于函数y=f(x),若数列{lnf(a n)}为等差数列,则称函数f(x)为“保比差数列函数”.现有定义在(0,+∞)上的如下函数:①f(x)=,②f(x)=x2,③f(x)=e x,④f(x)=,则为“保比差数列函数”的所有序号为( )(A)①②(B)③④(C)①②④(D)②③④二、填空题(本大题共5小题,每小题5分,共25分.请把正确答案填在题中横线上)11.已知数列{a n}的前n项和为S n=(-1)n n,则a n= .12.设{lga n}成等差数列,公差d=lg3,且{lga n}的前三项和为6lg3,则{a n}的通项公式为.13.已知函数f(x)对应关系如表所示,数列{a n}满足a1=3,a n+1=f(a n),则a2013= .14.(2013·咸阳模拟)设数列{a n}为等差数列,其前n项和为S n,a1+a4+a7=99,a2+a5+a8=93,若对任意n∈N+,都有S n≤S k成立,则正整数k的值为.15.(能力挑战题)已知数列{a n}的前n项和为S n,f(x)=,a n=log2,则S2013= .三、解答题(本大题共6小题,共75分.解答时应写出必要的文字说明、证明过程或演算步骤)16.(12分)(2013·宝鸡模拟)已知函数f(x)=log2x-x+1(x∈[2,+∞)),数列{a n}满足a 1=2,=2(n∈N+).(1)求数列{a n}的通项公式a n.(2)求f(a1)+f(a2)+…+f(a n).17.(12分)(2013·万州模拟)已知数列{a n}是首项a1=4,公比q≠1的等比数列,S n 是其前n项和,且4a1,a5,-2a3成等差数列.(1)求公比q的值.(2)设A n=S1+S2+S3+…+S n,求A n.18.(12分)已知数列{a n}中,a1=3,a n+1=2a n-1(n∈N+).(1)求证:数列{a n-1}是等比数列.(2)设b n=,求证:数列{b n}的前n项和S n<.19.(12分)某牛奶厂2009年初有资金1000万元,由于引进了先进设备,资金年平均增长率可达到50%.每年年底扣除下一年的消费基金x万元后,剩余资金投入再生产.(1)分别写出这家牛奶厂2010年初和2011年初投入再生产的剩余资金的表达式.(2)预计2013年底,这家牛奶厂将转向经营,需资金2000万元(该年底不再扣除下年的消费基金),当消费基金x不超过多少万元时,才能实现转向经营的目标(精确到万元)?20.(13分)(2012·山东高考)在等差数列{a n}中,a3+a4+a5=84,a9=73.(1)求数列{a n}的通项公式.(2)对任意m∈N+,将数列{a n}中落入区间(9m,92m)内的项的个数记为b m,求数列{b m}的前m项和S m.21.(14分)(能力挑战题)已知数列{a n}中a1=2,a n+1=2-,数列{b n}中b n=,其中n ∈N+.(1)求证:数列{b n}是等差数列.(2)设S n是数列{b n}的前n项和,求++…+.(3)设T n是数列{()n·b n}的前n项和,求证:T n<.答案解析1.【解析】选 C.设公差为d,则a1+d=3,2a1+5d=12,解得a1=1,d=2,所以a7+a8+a9=3a1+21d=3+42=45.2.【解析】选A.设公差为d,则S3=3a1+3d=6,即a1+d=2,所以5a1+a7=6a1+6d=12.3.【解析】选A.∵a4=2a3,S4=1,则q≠1,∴∴q=2,a1=,∴S8==17.4.【解析】选D.由+=3,得=3,又a2a3=a1a4=,则解得则q=2.所以a3+a4+a5+a6+a7+a8==63.5.【思路点拨】寻找数列的偶数项组成的数列的特点.【解析】选B.由题a n+1·a n=2n,a n+2·a n+1=2n+1,故=2,又a1=1,可得a2=2,故a10=25=32,选B.6.【解析】选D.由已知f(x+1)-f(x)=,得数列{f(n)}是等差数列,公差为,其前20项和为20×+×=335,故选D.7.【解析】选C.a3+a9+a15+a17=4a11=0,∴a11=0,S21=21a11=0.8.【解析】选B.∵-=2,∴-=2,故a12-a10=4,∴2d=4,d=2.∴S2012=2012a1+=-2012.9.【解析】选A.∵a n=+2(n-1),∴S n=na n-2n(n-1) ①∴S n+1=(n+1)a n+1-2(n+1)·n ②由②-①得:a n+1=(n+1)a n+1-na n-2n(n+1)+2n(n-1),化简得:na n+1-na n-4n=0,∴a n+1-a n=4,故数列{a n}是以a1=1为首项,d=4为公差的等差数列, a n=4n-3.∵S1+++…+-(n-1)2=2013,又∵=2n-1,∴1+3+5+…+(2n-1)-(n-1)2=2013,即-(n-1)2=2013⇒n=1007.10.【解析】选C.设数列{a n}的公比为q.①中,lnf(a n+1)-lnf(a n)=ln=ln=-lnq.故①中的函数符合要求;②中,lnf(a n+1)-lnf(a n)=ln=2lnq,也符合要求;③中,lnf(a n+1)-lnf(a n)=a n+1-a n,不符合要求;④中,lnf(a n+1)-lnf(a n)=ln=lnq,符合要求.11.【解析】当n≥2时,a n=S n-S n-1=(-1)n n-(-1)n-1(n-1)=(-1)n(2n-1),当n=1时也适合这个公式.答案：(-1)n(2n-1)12.【解析】根据等差数列性质可得lga2=2lg3,故数列{lga n}的通项公式是lga n=lga2+(n-2)lg3=nlg3=lg3n,所以a n=3n.答案：a n=3n13.【思路点拨】解答此类题目应先找规律,即先求a2,a3,a4,从中找出周期变化的规律.【解析】由题意知a2=f(a1)=f(3)=1,a3=f(a2)=f(1)=3,a4=f(a3)=f(3)=1,∴数列{a n}是周期为2的数列,∴a2013=a1=3.答案：314.【解析】方法一:由对任意n∈N+,都有S n≤S k成立,S k是S n的最大值.由等差数列的性质,有a1+a7=2a4,a2+a8=2a5,代入已知条件,得a4=33,a5=31,则公差d=a5-a4=-2,a1=33-3d=39,∴S n=39n+×(-2)=-n2+40n=-(n-20)2+400,则当n=20时,S n有最大值,故k的值为20.方法二:由题设对任意n∈N+,都有S n≤S k成立,求k的值即求S n最大时的项数n. 由等差数列的性质,有a1+a7=2a4,a2+a8=2a5,代入已知条件,得a4=33,a5=31,则公差d=a5-a4=-2,a1=33-3d=39,∴a n=39-2(n-1)=41-2n.由即解得20.5≥n>19.5,当n=20时,S n取得最大值,故k=20.答案：2015.【思路点拨】根据对数性质得a n=log2f(n+1)-log2f(n),裂项相消求和.【解析】由已知,得f(n)=,log2f(n)=log2,∴a n=log2=log2f(n+1)-log2f(n),∴S n=a1+a2+a3+…+a n=[log2f(2)-log2f(1)]+[log2f(3)-log2f(2)]+…+[log2f(n+1)-log2f(n)]=log2f(n+1)-log2f(1),则S2013=log2-log2=log2+1.答案:log2+116.【解析】(1)∵a 1=2,=2,∴{a n}是公比为2,首项为2的等比数列, ∴a n=2×2n-1=2n.(2)由(1)知f(a n)=log22n-2n+1=(n+1)-2n,则f(a1)+f(a2)+…+f(a n)=[2+3+…+(n+1)]-(2+22+…+2n)=-=-2n+1+2=n2+n+2-2n+1.17.【解析】(1)∵4a1,a5,-2a3成等差数列,∴2a5=4a1-2a3,∴2a1q4=4a1-2a1q2,∴q2=1,又q≠1,∴q=-1.(2)∵S n==2(1-(-1)n),∴A n=2(1-(-1)1)+2(1-(-1)2)+2(1-(-1)3)+…+2(1-(-1)n)=2(n-)=2n+1-(-1)n.18.【解析】(1)由a n+1=2a n-1,得a n+1-1=2(a n-1).即=2,∴数列{a n-1}是公比为2的等比数列.(2)由(1)知{a n-1}是公比为2,首项为2的等比数列,故a n-1=2n,∴a n=2n+1,∴b n====-∴S n=(-)+(-)+…+(-)=-<.【方法技巧】构造法求递推数列的通项公式对于由递推公式所确定的数列的求解,通常可通过对递推公式的变换转化,构造出等差数列或等比数列.一般根据递推式子的特点采取以下方法:(1)递推式为a n+1=qa n(q为常数):作商构造.(2)递推式为a n+1=a n+f(n):累加构造.(3)递推式为a n+1=pa n+q(p,q为常数):待定系数构造.(4)递推式为a n+1=pa n+q n(p,q为常数):辅助数列构造.(5)递推式为a n+2=pa n+1+qa n:待定系数构造.思路:设a n+2=pa n+1+qa n可以变形为:a n+2-αa n+1=β(a n+1-αa n),就是a n+2=(α+β)a n+1-αβa n,则可从解得α,β,于是{a n+1-αa n}是公比为β的等比数列,就转化为前面的类型.(6)递推式为a n+1=f(n)a n(n∈N+):累乘构造.(7)递推式为a n-a n-1+pa n a n-1=0(p为常数):倒数构造.【变式备选】已知数列{a n}满足:++…+=(32n-1),n∈N+.(1)求数列{a n}的通项公式.(2)设b n=log3,求++…+.【解析】(1)=(32-1)=3,当n≥2时,∵=(++…+)-(++…+)=(32n-1)-(32n-2-1)=32n-1,当n=1时,=32n-1也成立,∴数列{a n}的通项公式为a n=(n∈N+).(2)b n=log3=-(2n-1),==(-),∴++…+=[(1-)+(-)+…+(-)]=(1-)=.19.【解析】(1)2010年初的剩余资金为1000·-x;2011年初的剩余资金为(1000·-x)·-x.(2)设从2009年底这家牛奶厂的资金组成数列为{a n},则这个数列满足a1=1000·-x,a n+1=a n-x.设a n+1+λ=(a n+λ),展开与a n+1=a n-x比较可得λ=-2x,即a n+1=a n-x可以变换为a n+1-2x=(a n-2x),即数列{a n-2x}是首项为1000·-3x,公比为的等比数列,所以a n-2x=(1000·-3x)·()n-1,即a n=2x+(1000·-3x)·()n-1.从2009年初到2013年底共计5年,所以到2013年底该牛奶厂剩余资金a5=2x+(1000·-3x)·()4,只要a5+x≥2000,即2x+(1000·-3x)·()4+x≥2000即可,解得x≤≈458.97(万元).故当消费基金不超过458万元时,才能实现转向经营的目标.20.【思路点拨】(1)根据等差数列通项的性质求出a4,结合a9求出公差,进而得通项公式.(2)得出关于m,n的不等式,可得{b m}的通项公式,然后求和.【解析】(1)根据等差数列的性质得a4=28,设等差数列的公差为d,则a9-a4=5d=73-28=45,所以d=9,所以等差数列的通项公式为a n=a4+(n-4)d=28+(n-4)×9=9n-8,即a n=9n-8.(2)根据已知得9m<9n-8<92m,解得<n<,所以其中第一个n值为9m-1+1,最后一个n值为92m-1,所以b m=92m-1-9m-1,所以S m=(91-90)+(93-91)+…+(92m-1-9m-1)=(91+93+…+92m-1)-(90+91+…+9m-1)=-=-=.21.【解析】(1)b==,而b n=,∴b n+1-b n=-=1,n∈N+,∴{b n}是首项为b1==1,公差为1的等差数列.(2)由(1)可知b n=n,b n=n,∴S n=(1+2+…+n)=,于是==6(-),故有++…+=6(1-+-+…+-)=6(1-)=.(3)由(1)可知()n·b n=n·()n,则T n=1·+2·()2+…+n·()n,∴T n=1·()2+2·()3+…+(n-1)()n+n·()n+1.则T n=+()2+()3+…+()n-n()n+1=[1-()n]-n·()n+1,∴T n=-()n-1-·()n<.关闭Word文档返回原板块。

回夺市安然阳光实验学校课时提升作业五氧化还原反应(45分钟100分)一、选择题(本题包括12小题，每小题5分，共60分)1.(2016·宜春模拟)工业上生产下列物质，不涉及氧化还原反应的是( )A.用铝矾土(主要成分Al2O3)生产金属铝B.用硫铁矿(主要成分FeS2)生产硫酸C.用海水、贝壳生产氯化镁晶体D.用氯化钠生产烧碱【解析】选C。

Al2O3生产金属铝，铝元素化合价降低；FeS2生产硫酸，硫元素化合价升高；海水、贝壳生产氯化镁晶体时，CaCO3——CaO——Ca(OH)2——Mg(OH)2——MgCl2，元素化合价没有发生变化；氯化钠生产烧碱时生成氢气和氯气，元素化合价发生变化。

2.(2016·福州模拟)下表中对应关系正确的是( )A 向某溶液中加入盐酸产生无色气体溶液中一定含有C O32−B由油脂得到甘油由淀粉得到葡萄糖均发生了水解反应CCl2+2Br-2Cl-+Br2Zn+Cu2+Zn2++Cu均为单质被还原的置换反应D2Na2O2+2H2O4NaOH+O2↑Cl2+H2O HCl+HClO均为水作还原剂的氧化还原反应【解析】选B。

无色气体可能为二氧化碳或者二氧化硫，A错误；油脂为高级脂肪酸甘油酯，水解生成甘油；淀粉为多糖，水解最终产物为葡萄糖，则均可发生水解反应，B正确；Cl2+2Br-2Cl-+Br2中Cl的化合价降低，单质被还原；Zn+Cu2+Zn2++Cu中Zn的化合价升高，单质被氧化，均属于置换反应，C错误；前者只有过氧化钠中氧元素的化合价变化，水既不是氧化剂也不是还原剂；Cl2+H2O HCl+HClO中只有Cl的化合价变化，均属于氧化还原反应，但水既不是氧化剂也不是还原剂，D错误。

3.(2016·宣城模拟)有人说“五颜六色”形象地说出了化学实验中的颜色变化。

下列颜色变化中是由于发生氧化还原反应导致的是( )①在氯水中加入NaOH溶液；②在FeCl3溶液中加入铁粉；③在品红溶液中通入二氧化硫气体；④在Na2CO3溶液中滴入酚酞；⑤在新制Cu(OH)2悬浊液中滴入葡萄糖溶液，加热。

课时提升作业(五)(45分钟100分)一、选择题(本大题共12小题，每小题5分，共60分)1.(2013·南通模拟)美国历史学家费正清在《伟大的中国革命》一书中表达了这样的观点：“鸦片战争的发生意味着中国拒绝在外交平等和对等贸易的基础上参加国际大家庭，结果导致英国使用武力。

”费正清的上述观点( )A.解释了鸦片战争的根源B.肯定了中国抗击英国侵略的正义性C.抹杀了鸦片战争的实质D.从全球化角度正确分析了战争起因2.(2013·宿迁一模)有学者认为：“就世界大势论，鸦片战争是不能避免的。

”这里所说的“大势”主要是指( )A.英国的船坚炮利B.工业文明迅速发展C.中国的闭关锁国D.西方列强加紧扩张3.《南京条约》签订后，清政府并未在国内予以公布颁行，当时连清朝办理“夷务”的官员都“未见其文”，社会各界对条约具体内容更是缺乏了解。

第二次鸦片战争后，咸丰帝颁发上谕，准允将《天津条约》《北京条约》印刷成册，颁发各省。

这说明( )A.第一个不平等条约对中国社会影响有限B.闭关锁国政策对不平等条约有抵制作用C.清政府对条约从不接受到逐渐认同D.列强的侵略促进了清政府政治公正透明4.1852年英国驻广州代办密切尔惊异地说：“经过和这么一个大国开放贸易10年之久，并且双方都已废除了一切独立制度。

而拥有如此庞大人口的中国，其消费我们的制造品竟不及荷兰的一半……这好像是一个奇怪的结局。

”在当时的英国人看来，这一现象出现的最主要原因是( )A.中国自然经济的阻碍B.中国没能给予最惠国待遇C.中国人消费能力较低D.中国的对外通商口岸太少5.《字林沪报》是中国近代有较大影响力的报刊之一。

有人在这份报刊上查阅到这样一段文字：“高丽之乱国固非自今日始矣。

光绪八年壬午岁，大院君之变，则日本暗主其谋。

光绪十年甲申岁，金玉均之叛，则日本明助其力。

若今东学党之事，不过乱民匪党，癣疥之疾，而竟兵连祸结，若临大敌者，此中未必无人焉”。