章末检测试卷一(第四章).pptx

Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,11/7/2009,#,章末检测试卷一,(,第四章,),第四章,数,列,第一页，编辑于星期五：十四点 二十四分。,A.,第,10,项,B.,第,11,项,C,.,第,12,项,D.,第,21,项,一、单项选择题,(,本大题共,8,小题，每小题,5,分，共,40,分,),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,解析,观察可知该数列的通项公式为,a,n,(,事实上，根号内的数成等差数列，首项为,1,，公差为,2),，,令,21,2,n,1,，解得,n,11.,第二页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,解析,a,1,a,5,2,a,3,10,，,a,3,5,，,d,a,4,a,3,7,5,2.,2.,在等差数列,a,n,中，,a,1,a,5,10,，,a,4,7,，则数列,a,n,的公差为,A.1,B.2 C.3 D.4,20,第三页，编辑于星期五：十四点 二十四分。,3.,在等差数列,a,n,中，若,a,2,a,3,4,，,a,4,a,5,6,，则,a,9,a,10,等于,A.9,B.10 C.11 D.12,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,解析,设等差数列,a,n,的公差为,d,，,20,又,(,a,9,a,10,),(,a,4,a,5,),10,d,5,，,所以,a,9,a,10,(,a,4,a,5,),5,11.,第四页，编辑于星期五：十四点 二十四分。,4.,设等差数列,a,n,的前,n,项和是,S,n,，若,a,m,a,1,0,，且,S,m,1,0,B.,S,m,0,C.,S,m,0且,S,m,1,0,D.,S,m,0,，且,S,m,1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,解析,因为,a,m,a,1,0,，,a,1,a,m,1,0,，且,S,m,1,0.,20,第五页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,5.,设,S,n,为等比数列,a,n,的前,n,项和，,a,1,1,且,a,1,a,2,a,3,8,，,则,等于,A.,11,B,.,8,C.5 D.11,解析,设等比数列,a,n,的公比为,q,，,又,a,1,1,，,第六页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,6.,已知公差不为,0,的等差数列,a,n,的前,23,项的和等于前,8,项的和,.,若,a,8,a,k,0,，则,k,等于,A.22,B.23 C.24 D.25,解析,等差数列的前,n,项和,S,n,可看作关于,n,的二次函数,(,图象过原点,).,所以,S,15,S,16,，即,a,16,0,，,所以,a,8,a,24,2,a,16,0,，所以,k,24.,第七页，编辑于星期五：十四点 二十四分。1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,7.,风雨桥是侗族最具特色的建筑之一,.,风雨桥由桥、塔、亭组成,.,其亭、塔平面图通常是正方形、正六边形和正八边形,.,如图是风雨桥亭、塔正六边形的正射影,.,其正六边形的边长计算方法如下：,A,1,B,1,A,0,B,0,B,0,B,1,，,A,2,B,2,A,1,B,1,B,1,B,2,，,A,3,B,3,A,2,B,2,B,2,B,3,，,，,A,n,B,n,A,n,1,B,n,1,B,n,1,B,n,，其中,B,n,1,B,n,B,2,B,3,B,1,B,2,B,0,B1,，,n,N,*,.,根据每层边长间的规律,.,建筑师通过推算，可初步估计需要多少材料,.,所用材料中,.,横向梁所用木料与正六边形的周长有关,.,某一风雨桥亭、塔共,5,层，若,A,0,B,0,8 m,，,B,0,B,1,0.5 m,.,则,这五层正六边形的周长总和为,A.35 m,B.45,m,C.210 m,D.270,m,第八页，编辑于星期五：十四点 二十四分。,解析,由已知得：,A,n,B,n,A,n,1,B,n,1,B,n,1,B,n,，,B,n,1,B,n,B,2,B,3,B,1,B,2,B,0,B,1,0.5,，,因此数列,A,n,B,n,(,n,N,*,1,n,5),是以a,1,A,0,B,0,8,为首项，公差为,d,0.5,的等差数列,，,设,数列,A,n,B,n,(,n,N,*,1,n,5),前,5,项和为,S,5,，,所以这五层正六边形的周长总和为,6,S,5,6,35,210 m.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第九页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,A.220,B.110 C.99 D.55,20,所以,a,n,an,2,bn,.,因为,a,1,2,4,a,3,a,6,，所以,a,b,2,，且,4(9,a,3,b,),36,a,6,b,，,解得,a,2,，,b,0,，所以,a,n,2,n,2,.,所以,S,10,2(,1,2,2,2,),(,3,2,4,2,),(,9,2,10,2,),110.,第十页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,二、多项选择题,(,本大题共,4,小题，每小题,5,分，共,20,分,.,全部选对的得,5,分，部分选对的得,3,分，有选错的得,0,分,),9.,已知数列,a,n,的通项公式为,a,n,9,2,n,，则下列各数中是,a,n,中的项的是,A.0,B.3 C.5 D.7,20,对于,B,3,9,2,n,，解得,n,3,，故,B,满足；,对于,C,5,9,2,n,，解得,n,2,，故,C,满足；,对于,D,7,9,2,n,，解得,n,1,，故,D,满足,.,第十一页，编辑于星期五：十四点 二十四分。,10.,在等比数列,a,n,中，已知,a,1,3,，,a,3,27,，则数列的通项公式是,A.,a,n,3,n,，,n,N,*,B.,a,n,3,n,1,，,n,N,*,C.,a,n,(,1),n,1,3,n,，,n,N,*,D.,a,n,2,n,1,，,n,N,*,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,解析,由,a,3,a,1,q,2,，得,q,2,9,，即,q,3.,a,n,a,1,q,n,1,3,3,n,1,3,n,或,a,n,a,1,q,n,1,3,(,3),n,1,(,1),n,1,3,n,.,故,数列的通项公式是,a,n,3,n,(,n,N,*,),或,a,n,(,1),n,1,3,n,，,n,N,*,.,第十二页，编辑于星期五：十四点 二十四分。,A.2,B.3 C.4 D.14,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第十三页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,则,n,1,的可能取值有,3,5,15,，,因此，正整数,n,的可能取值有,2,4,14.,第十四页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,A.,数列,a,n,1,是等差数列,B.,数列,a,n,1,是等比数列,C.,数列,a,n,的通项公式为,a,n,2,n,1,D.,T,n,1.,求证：,(1),S,n,2,a,n,1,；,20,第二十七页，编辑于星期五：十四点 二十四分。,证明,因为数列,a,n,为等比数列,，所以,a,1,a,6,a,3,a,4,32.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第二十八页，编辑于星期五：十四点 二十四分。,证明,由,(1),知,a,n,2,n,1,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第二十九页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,20.(12,分,),在,a,3,5,，,a,2,a,5,6,b,2,；,b,2,2,，,a,3,a,4,3,b,3,；,S,3,9,，,a,4,a,5,8,b,2,，这三个条件中任选一个，补充在下面问题中，并解答,.,已知等差数列,a,n,的公差为,d,(,d,1),，前,n,项和为,S,n,，等比数列,b,n,的公比为,q,，且,a,1,b,1,，,d,q,，,.,(1),求数列,a,n,，,b,n,的通项公式；,第三十页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,解,方案一：选条件,.,(1),a,3,5,，,a,2,a,5,6,b,2,，,a,1,b,1,，,d,q,，,d,1,，,20,a,n,a,1,(,n,1),d,2,n,1,，,n,N,*,，,b,n,b,1,q,n,1,2,n,1,，,n,N,*,.,第三十一页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第三十二页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,方案二：选条件,.,(1),b,2,2,，,a,3,a,4,3,b,3,，,a,1,b,1,，,d,q,，,d,1,，,a,n,a,1,(,n,1),d,2,n,1,，,n,N,*,，,b,n,b,1,q,n,1,2,n,1,，,n,N,*,.,第三十三页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第三十四页，编辑于星期五：十四点 二十四分。,方案三：选条件,.,(1),S,3,9,，,a,4,a,5,8,b,2,，,a,1,b,1,，,d,q,，,d,1,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第三十五页，编辑于星期五：十四点 二十四分。,a,n,a,1,(,n,1),d,2,n,1,，,n,N,*,，,b,n,b,1,q,n,1,2,n,1,，,n,N,*,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第三十六页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第三十七页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,21.(12,分,),某市,2018,年发放汽车牌照,12,万张，其中燃油型汽车牌照,10,万张，电动型汽车牌照,2,万张,.,为了节能减排和控制汽车总量，从,2018,年开始，每年电动型汽车牌照按,50%,增长，而燃油型汽车牌照每一年比上一年减少,0.5,万张，同时规定一旦某年发放的牌照超过,15,万张，以后每一年发放的电动型的牌照的数量维持在这一年的水平不变,.,(1),记,2018,年为第一年，每年发放的燃油型汽车牌照数构成数列,a,n,，每年发放的电动型汽车牌照数构成数列,b,n,，完成下列表格，并写出这两个数列的通项公式,.,20,a,1,10,a,2,9.5,a,3,a,4,b,1,2,b,2,b,3,b,4,第三十八页，编辑于星期五：十四点 二十四分。,解,a,1,10,a,2,9.5,a,3,9,a,4,8.5,b,1,2,b,2,3,b,3,4.5,b,4,6.75,当,1,n,20,且,n,N,*,时，,a,n,10,(,n,1),(,0.5),0.5,n,10.5,；,当,n,21,且,n,N,*,时，,a,n,0.,而,a,4,b,4,15.2515,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第三十九页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,(2),从,2018,年算起，累计各年发放的牌照数，哪一年开始超过,200,万张？,20,第四十页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,解,当,n,4,时，,S,n,a,1,a,2,a,3,a,4,b,1,b,2,b,3,b,4,53.25.,当,5,n,21,时，,S,n,(,a,1,a,2,a,n,),(,b,1,b,2,b,3,b,4,b,5,b,n,),所以结合实际情况，可知到,2034,年累计发放汽车牌照超过,200,万张,.,第四十一页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,22.(12,分,),在等差数列,a,n,中，,a,3,6,，,a,8,26,，,S,n,为等比数列,b,n,的前,n,项和，且,b,1,1,4,S,1,3,S,2,2,S,3,成等差数列,.,(,1),求数列,a,n,与,b,n,的通项公式；,20,第四十二页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,解,设等差数列,a,n,的公差为,d,，,等比数列,b,n,的公比为,q,.,由题意，得,a,8,a,3,5,d,26,6,20,，,所以,d,4,，所以,a,n,a,3,(,n,3),d,4,n,6,，,n,N,*,.,因为,6,S,2,4,S,1,2,S,3,，即,3(,b,1,b,2,),2,b,1,b,1,b,2,b,3,，,所以,b,3,2,b,2,.,所以公比,q,2,，所以,b,n,2,n,1,，,n,N,*,.,20,第四十三页，编辑于星期五：十四点 二十四分。,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,222),设,c,n,|,a,n,|,b,n,，求数列,c,n,的前,n,项和,T,n,.,20,第四十四页，编辑于星期五：十四点 二十四分。,解,由,(1),可得，,c,n,|4,n,6|2,n,1,|2,n,3|2,n,.,当,n,1,时，,2,n,30,，所以,c,n,(2,n,3)2,n,，,T,n,2,1,2,2,3,2,3,5,2,4,(2,n,3),2,n,，,所以,2,T,n,4,1,2,3,3,2,4,(2,n,3),2,n,1,.,所以,得，,T,n,2,2,(2,3,2,4,2,n,),(2,n,3),2,n,1,所以,T,n,(2,n,5)2,n,1,14,.,当,n,1,时，满足上式,.,所以,T,n,(2,n,5)2,n,1,14,，,n,N,*,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,第四十五页，编辑于星期五：十四点 二十四分。,本课结束,更多精彩内容请登录,：,第四十六页，编辑于星期五：十四点 二十四分。,