高考数学专题三-第1讲-等差数列与等比数列课件

-,-,核心知识,整合,第,1,讲等差数列与等比数列,高频考点,突破,高考真题,诊断,第,1,讲等差数列与等比数列,近五年高考试题统计与命题,预测,1,.,(,2019,全国,文,6,),已知各项均为正数的等比数列,a,n,的前,4,项和为,15,且,a,5,=,3,a,3,+,4,a,1,则,a,3,=,(,),A.16B.8C.4D.2,解析,:,设等比数列,a,n,的公比为,q,(,q,0),所以,a,3,=a,1,q,2,=,1,2,2,=,4,.,故选,C,.,答案,:,C,答案,:,A,4,.,(2019,全国,文,14),记,S,n,为等差数列,a,n,的前,n,项和,.,若,a,3,=,5,a,7,=,13,则,S,10,=,.,答案,:,100,二、等差、等比数列的判定与证明,证明数列,a,n,是等差数列或等比数列的方法,(1),证明数列,a,n,是等差数列的两种基本方法,:,利用定义,证明,a,n+,1,-a,n,(,n,N,*,),为一常数,;,利用等差中项,即证明,2,a,n,=a,n-,1,+a,n+,1,(,n,2),.,(2),证明,a,n,是等比数列的两种基本方法,:,考点,1,考点,2,考点,3,等差、等比数列基本运算,(,基本元思想,),例,1,(1)(,2019,天津和平区质检,),已知等比数列,a,n,满足,a,1,=,1,a,3,a,5,=,4(,a,4,-,1),则,a,7,的值为,(,),A.2B.4,C,.,D.6,(2)(,2018,全国,文,17,),记,S,n,为等差数列,a,n,的前,n,项和,已知,a,1,=-,7,S,3,=-,15,.,求,a,n,的通项公式,;,求,S,n,并求,S,n,的最小值,.,考点,1,考点,2,考点,3,(1),解析,:,根据等比数列的性质,得,a,3,a,5,=,=,4(,a,4,-,1),即,(,a,4,-,2),2,=,0,解得,a,4,=,2,.,又,a,1,=,1,a,1,a,7,=,4,a,7,=,4,.,答案,:,B,(2),解,:,设,a,n,的公差为,d,由题意得,3,a,1,+,3,d=-,15,.,由,a,1,=-,7,得,d=,2,.,所以,a,n,的通项公式为,a,n,=,2,n-,9,.,由,得,S,n,=n,2,-,8,n=,(,n-,4),2,-,16,.,所以当,n=,4,时,S,n,取得最小值,最小值为,-,16,.,考点,1,考点,2,考点,3,考点,1,考点,2,考点,3,对应训练,1,(1)(2019,山东潍坊检测,),设等差数列,a,n,的前,n,项和为,S,n,S,11,=,22,a,4,=-,12,若,a,m,=,30,则,m=,(,),A.9B.10C.11D.15,(,3)(2018,全国,文,17),等比数列,a,n,中,a,1,=,1,a,5,=,4,a,3,.,求,a,n,的通项公式,;,记,S,n,为,a,n,的前,n,项和,若,S,m,=,63,求,m.,考点,1,考点,2,考点,3,考点,1,考点,2,考点,3,(3),解,:,设,a,n,的公比为,q,由题设得,a,n,=q,n-,1,.,由已知得,q,4,=,4,q,2,解得,q=,0(,舍去,),q=-,2,或,q=,2,.,故,a,n,=,(,-,2),n-,1,或,a,n,=,2,n-,1,.,由,S,m,=,63,得,(,-,2),m,=-,188,此方程没有正整数解,.,若,a,n,=,2,n-,1,则,S,n,=,2,n,-,1,.,由,S,m,=,63,得,2,m,=,64,解得,m=,6,.,综上,m=,6,.,考点,1,考点,2,考点,3,等差、等比数列的判定与证明,例,2,(1),记,S,n,为数列,a,n,的前,n,项和,.,若,S,n,=,2,a,n,+,1,则,S,6,=,.,(2)(2018,全国,文,17),已知数列,a,n,满足,a,1,=,1,na,n+,1,=,2(,n+,1),a,n,.,设,求,b,1,b,2,b,3,;,判断数列,b,n,是否为等比数列,并说明理由,;,求,a,n,的通项公式,.,(3)(2019,广东省级名校联考,),已知,S,n,是数列,a,n,的前,n,项和,且满足,S,n,-,2,a,n,=n-,4,.,证明,:,S,n,-n+,2,为等比数列,;,求数列,S,n,的前,n,项和,T,n,.,考点,1,考点,2,考点,3,(1),解析,:,S,n,=,2,a,n,+,1,S,n-,1,=,2,a,n-,1,+,1(,n,2),.,-,得,a,n,=,2,a,n,-,2,a,n-,1,即,a,n,=,2,a,n-,1,(,n,2),.,又,S,1,=,2,a,1,+,1,a,1,=-,1,.,答案,:,-,63,考点,1,考点,2,考点,3,考点,1,考点,2,考点,3,考点,1,考点,2,考点,3,考点,1,考点,2,考点,3,对应训练,2,(,1,如,图,点列,A,n,B,n,分别在某锐角的两边上,且,|A,n,A,n+,1,|=|A,n+,1,A,n+,2,|,A,n,A,n+,2,n,N,*,|B,n,B,n+,1,|=|B,n+,1,B,n+,2,|,B,n,B,n+,2,n,N,*,.,(,P,Q,表示点,P,与,Q,不重合,),若,d,n,=|A,n,B,n,|,S,n,为,A,n,B,n,B,n+,1,的面积,则,(,),(,2,),已知,数列,a,n,的前,n,项和,S,n,=,1,+,a,n,其中,0,.,证明,a,n,是等比数列,并求其通项公式,;,考点,1,考点,2,考点,3,(1),解析,:,如图,延长,A,n,A,1,B,n,B,1,交于,P,过,A,n,作对边,B,n,B,n+,1,的垂线,其长度记为,h,1,过,A,n+,1,作对边,B,n+,1,B,n+,2,的垂线,其长度记为,h,2,设,此锐角为,则,h,2,=|PA,n+,1,|,sin,h,1,=|PA,n,|,sin,h,2,-h,1,=,sin,(,|PA,n+,1,|-|PA,n,|,),=|A,n,A,n+,1,|,sin,.,S,n+,1,-S,n,=|,B,n,B,n+,1,|A,n,A,n+,1,|,sin,.,|B,n,B,n+,1,|,|A,n,A,n+,1,|,sin,均为定值,S,n+,1,-S,n,为定值,.,S,n,是等差数列,.,故选,A,.,答案,:,A,考点,1,考点,2,考点,3,考点,1,考点,2,考点,3,等差、等比数列综合、创新题型,例,3,(1,)(2018,浙江,10),已知,a,1,a,2,a,3,a,4,成等比数列,且,a,1,+a,2,+a,3,+a,4,=,ln(,a,1,+a,2,+a,3,),.,若,a,1,1,则,(,),A,.a,1,a,3,a,2,a,3,a,2,a,4,C,.a,1,a,4,D,.a,1,a,3,a,2,a,4,(,2)(2018,上海,21),给定无穷数列,a,n,若无穷数列,b,n,满足,:,对任意,x,N,*,都有,|b,n,-a,n,|,1,则称,b,n,与,a,n,“,接近,”,.,设,a,n,是首项为,1,公比,为,的,等比数列,b,n,=a,n+,1,+,1,n,N,*,判断数列,b,n,是否与,a,n,接近,并说明理由,;,设数列,a,n,的前四项为,a,1,=,1,a,2,=,2,a,3,=,4,a,4,=,8,b,n,是一个与,a,n,接近的数列,记集合,M=,x|x=b,i,i=,1,2,3,4,求,M,中元素的个数,m,:,已知,a,n,是公差为,d,的等差数列,.,若存在数列,b,n,满足,:,b,n,与,a,n,接近,且在,b,2,-b,1,b,3,-b,2,b,201,-b,200,中至少有,100,个为正数,求,d,的取值范围,.,考点,1,考点,2,考点,3,考点,1,考点,2,考点,3,(1),解析,:,设等比数列的公比为,q,a,1,+a,2,+a,3,+a,4,=,ln(,a,1,+a,2,+a,3,),又,a,1,1,q,1,即,q+q,2,0,解得,q,0,舍去,),.,由,a,1,1,可知,a,1,(1,+q+q,2,),1,a,1,(1,+q+q,2,+q,3,),0,即,1,+q+q,2,+q,3,0,即,(1,+q,),+q,2,(1,+q,),0,即,(1,+q,)(1,+q,2,),0,这与,q-,1,相矛盾,.,1,+q+q,2,1,即,-,1,qa,3,a,2,0,取,b,n,=a,n,可得,b,n+,1,-b,n,=a,n+,1,-a,n,=d,0,则,b,2,-b,1,b,3,-b,2,b,201,-b,200,中有,200,个正数,符合题意,;,则,b,2,-b,1,b,3,-b,2,b,201,-b,200,中有,200,个正数,符合题意,;,(,),若,-,2,d,0,则,b,2,-b,1,b,3,-b,2,b,201,-b,200,中恰有,100,个正数,符合题意,;,考点,1,考点,2,考点,3,(,),若,d,-,2,假设存在数列,b,n,满足,:,b,n,与,a,n,接近,则为,a,n,-,1,b,n,a,n,+,1,a,n+,1,-,1,b,n+,1,a,n+,1,+,1,可得,b,n+,1,-b,n,a,n+,1,+,1,-,(,a,n,-,1),=,2,+d,0,b,2,-b,1,b,3,-b,2,b,201,-b,200,中没有正数,与已知矛盾,.,故,d,-,2,不符合题意,.,综上可得,d,的取值范围是,(,-,2,+,),.,考点,1,考点,2,考点,3,考点,1,考点,2,考点,3,对应训练,3,(1)(2017,浙江,6),已知等差数列,a,n,的公差为,d,前,n,项和为,S,n,则,“,d,0,”,是,“,S,4,+S,6,2,S,5,”,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,(2,),设,a,n,和,b,n,是两个等差数列,记,c,n,=,max,b,1,-a,1,n,b,2,-a,2,n,b,n,-a,n,n,(,n=,1,2,3,),其中,max,x,1,x,2,x,s,表示,x,1,x,2,x,s,这,s,个数中最大的数,.,若,a,n,=n,b,n,=,2,n-,1,求,c,1,c,2,c,3,的值,并证明,c,n,是等差数列,;,证明,:,或者对任意正数,M,存在正整数,m,当,n,m,时,M,;,或者存在正整数,m,使得,c,m,c,m+,1,c,m+,2,是等差数列,.,考点,1,考点,2,考点,3,(3)(2016,上海,理,23),若无穷数列,a,n,满足,:,只要,a,p,=a,q,(,p,q,N,*,),必有,a,p+,1,=a,q+,1,则称,a,n,具有性质,P.,若,a,n,具有性质,P,且,a,1,=,1,a,2,=,2,a,4,=,3,a,5,=,2,a,6,+a,7,+a,8,=,21,求,a,3,;,若无穷数列,b,n,是等差数列,无穷数列,c,n,是公比为正数的等比数列,b,1,=c,5,=,1,b,5,=c,1,=,81,a,n,=b,n,+c,n,判断,a,n,是否具有性质,P,并说明理由,;,设,b,n,是无穷数列,已知,a,n+,1,=b,n,+,sin,a,n,(,n,N,*,),求证,:,“,对任意,a,1,a,n,都具有性质,P,”,的充要条件为,“,b,n,是常数列,”,.,考点,1,考点,2,考点,3,所以,S,4,+S,6,2,S,5,10,a,1,+,21,d,10,a,1,+,20,d,d,0,即,“,d,0,”,是,“,S,4,+S,6,2,S,5,”,的充分必要条件,选,C,.,答案,:,C,(2),解,:,c,1,=b,1,-a,1,=,1,