2019届江苏专用高考数学一轮复习第六章数列6.2等差数列讲义.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,高考数学,(江苏省专用),6.2等差数列,1.,(2016江苏,8,5分)已知,a,n,是等差数列,S,n,是其前,n,项和.若,a,1,+,=-3,S,5,=10,则,a,9,的值是,.,A,组 自主命题,江苏卷题组,五年高考,答案,20,解析,设等差数列,a,n,的公差为,d,则由题设可得,解得,从而,a,9,=,a,1,+8,d,=20.,2.,(2017江苏,19,16分)对于给定的正整数,k,若数列,a,n,满足:,a,n,-,k,+,a,n,-,k,+1,+,+,a,n,-1,+,a,n,+1,+,+,a,n,+,k,-1,+,a,n,+,k,=2,ka,n,对任意正整数,n,(,n,k,)总成立,则称数列,a,n,是“,P,(,k,)数列”.,(1)证明:等差数列,a,n,是“,P,(3)数列”;,(2)若数列,a,n,既是“,P,(2)数列”,又是“,P,(3)数列”,证明:,a,n,是等差数列.,证明,本小题主要考查等差数列的定义、通项公式等基础知识,考查代数推理、转化与化归及,综合运用数学知识探究与解决问题的能力.,(1)因为,a,n,是等差数列,设其公差为,d,则,a,n,=,a,1,+(,n,-1),d,从而,当,n,4时,a,n,-,k,+,a,n,+,k,=,a,1,+(,n,-,k,-1),d,+,a,1,+(,n,+,k,-1),d,=2,a,1,+2(,n,-1),d,=2,a,n,k,=1,2,3,所以,a,n,-3,+,a,n,-2,+,a,n,-1,+,a,n,+1,+,a,n,+2,+,a,n,+3,=6,a,n,因此等差数列,a,n,是“,P,(3)数列”.,(2)数列,a,n,既是“,P,(2)数列”,又是“,P,(3)数列”,因此,当,n,3时,a,n,-2,+,a,n,-1,+,a,n,+1,+,a,n,+2,=4,a,n,当,n,4时,a,n,-3,+,a,n,-2,+,a,n,-1,+,a,n,+1,+,a,n,+2,+,a,n,+3,=6,a,n,.,由知,a,n,-3,+,a,n,-2,=4,a,n,-1,-(,a,n,+,a,n,+1,),a,n,+2,+,a,n,+3,=4,a,n,+1,-(,a,n,-1,+,a,n,).,将代入,得,a,n,-1,+,a,n,+1,=2,a,n,其中,n,4,所以,a,3,a,4,a,5,是等差数列,设其公差为,d,.,在中,取,n,=4,则,a,2,+,a,3,+,a,5,+,a,6,=4,a,4,所以,a,2,=,a,3,-,d,在中,取,n,=3,则,a,1,+,a,2,+,a,4,+,a,5,=4,a,3,所以,a,1,=,a,3,-2,d,所以数列,a,n,是等差数列.,方法总结,数列新定义型创新题的一般解题思路:,1.阅读审清“新定义”;,2.结合常规的等差数列、等比数列的相关知识,化归、转化到“新定义”的相关知识;,3.利用“新定义”及常规的数列知识,求解证明相关结论.,3.,(2013江苏,19,16分,0.224)设,a,n,是首项为,a,公差为,d,的等差数列(,d,0),S,n,是其前,n,项的和.记,b,n,=,n,N,*,其中,c,为实数.,(1)若,c,=0,且,b,1,b,2,b,4,成等比数列,证明:,S,nk,=,n,2,S,k,(,k,n,N,*,);,(2)若,b,n,是等差数列,证明:,c,=0.,证明,由题意得,S,n,=,na,+,d,.,(1)由,c,=0,得,b,n,=,=,a,+,d,.,又因为,b,1,b,2,b,4,成等比数列,所以,=,b,1,b,4,即,=,a,化简得,d,2,-2,ad,=0.,因为,d,0,所以,d,=2,a,.,因此,对于所有的,m,N,*,有,S,m,=,m,2,a,.,从而对于所有的,k,n,N,*,有,S,nk,=(,nk,),2,a,=,n,2,k,2,a,=,n,2,S,k,.,(2)设数列,b,n,的公差是,d,1,则,b,n,=,b,1,+(,n,-1),d,1,即,=,b,1,+(,n,-1),d,1,n,N,*,代入,S,n,的表达式,整理得,对,于所有的,n,N,*,有,n,3,+,n,2,+,cd,1,n,=,c,(,d,1,-,b,1,).,令,A,=,d,1,-,d,B,=,b,1,-,d,1,-,a,+,d,D,=,c,(,d,1,-,b,1,),则对于所有的,n,N,*,有,An,3,+,Bn,2,+,cd,1,n,=,D,.,(*),在(*)式中分别取,n,=1,2,3,4,得,A,+,B,+,cd,1,=8,A,+4,B,+2,cd,1,=27,A,+9,B,+3,cd,1,=64,A,+16,B,+4,cd,1,从而有,由,得,A,=0,cd,1,=-5,B,代入方程,得,B,=0,从而,cd,1,=0.,即,d,1,-,d,=0,b,1,-,d,1,-,a,+,d,=0,cd,1,=0.,若,d,1,=0,则由,d,1,-,d,=0,得,d,=0,与题设矛盾,所以,d,1,0.,又因为,cd,1,=0,所以,c,=0.,4.,(2014江苏,20,16分,0.17)设数列,a,n,的前,n,项和为,S,n,.若对任意的正整数,n,总存在正整数,m,使得,S,n,=,a,m,则称,a,n,是“,H,数列”.,(1)若数列,a,n,的前,n,项和,S,n,=2,n,(,n,N,*,),证明:,a,n,是“,H,数列”;,(2)设,a,n,是等差数列,其首项,a,1,=1,公差,d,0.若,a,n,是“,H,数列”,求,d,的值;,(3)证明:对任意的等差数列,a,n,总存在两个“,H,数列”,b,n,和,c,n,使得,a,n,=,b,n,+,c,n,(,n,N,*,)成立.,解析,(1)证明:由已知得,当,n,1时,a,n,+1,=,S,n,+1,-,S,n,=2,n,+1,-2,n,=2,n,.于是对任意的正整数,n,总存在正整数,m,=,n,+1,使得,S,n,=2,n,=,a,m,.,所以,a,n,是“,H,数列”.,(2)由已知,得,S,2,=2,a,1,+,d,=2+,d,.因为,a,n,是“,H,数列”,所以存在正整数,m,使得,S,2,=,a,m,即2+,d,=1+(,m,-1),d,于是(,m,-2),d,=1.因为,d,0,所以,m,-20,故,m,=1.从而,d,=-1.,当,d,=-1时,a,n,=2-,n,S,n,=,是小于2的整数,n,N,*,.于是对任意的正整数,n,总存在正整数,m,=2-,S,n,=,2-,使得,S,n,=2-,m,=,a,m,所以,a,n,是“,H,数列”.,因此,d,的值为-1.,(3)证明:设等差数列,a,n,的公差为,d,则,a,n,=,a,1,+(,n,-1),d,=,na,1,+(,n,-1)(,d,-,a,1,)(,n,N,*,).,令,b,n,=,na,1,c,n,=(,n,-1)(,d,-,a,1,),则,a,n,=,b,n,+,c,n,(,n,N,*,),下证,b,n,是“,H,数列”.,设,b,n,的前,n,项和为,T,n,则,T,n,=,a,1,(,n,N,*,).于是对任意的正整数,n,总存在正整数,m,=,使,得,T,n,=,b,m,.所以,b,n,是“,H,数列”.,同理可证,c,n,也是“,H,数列”.,所以,对任意的等差数列,a,n,总存在两个“,H,数列”,b,n,和,c,n,使得,a,n,=,b,n,+,c,n,(,n,N,*,).,考点一等差数列的定义及运算,1.,(2017课标全国理改编,9,5分)等差数列,a,n,的首项为1,公差不为0.若,a,2,a,3,a,6,成等比数列,则,a,n,前6项的和为,.,B组,统一命题,省(区、市)卷题组,答案,-24,解析,本题主要考查等差数列的通项公式及前,n,项和公式.,设等差数列,a,n,的公差为,d,依题意得,=,a,2,a,6,即(1+2,d,),2,=(1+,d,)(1+5,d,),解得,d,=-2或,d,=0(舍去),又,a,1,=,1,S,6,=6,1+,(-2)=-24.,2.,(2017课标全国理改编,4,5分)记,S,n,为等差数列,a,n,的前,n,项和.若,a,4,+,a,5,=24,S,6,=48,则,a,n,的公差,为,.,答案,4,解析,本题考查等差数列基本量的计算与性质的综合应用.,等差数列,a,n,中,S,6,=,=48,则,a,1,+,a,6,=16=,a,2,+,a,5,又,a,4,+,a,5,=24,所以,a,4,-,a,2,=2,d,=24-16=8,得,d,=4.,方法总结,求解此类题时,常用,S,n,=,求出,a,1,+,a,n,的值,再结合等差数列中“若,m,n,p,q,N,*,m,+,n,=,p,+,q,则,a,m,+,a,n,=,a,p,+,a,q,”的性质求解数列中的基本量.,3.,(2017课标全国理,15,5分)等差数列,a,n,的前,n,项和为,S,n,a,3,=3,S,4,=10,则,=,.,答案,解析,本题主要考查等差数列基本量的计算及裂项相消法求和.,设公差为,d,则,a,n,=,n,.,前,n,项和,S,n,=1+2+,+,n,=,=,=2,=2,1-,+,-,+,+,-,=2,=2,=,.,思路分析,求出首项,a,1,和公差,d,从而求出,S,n,.,=,=2,从而运用裂项相消法求和,即可.,解后反思,裂项相消法求和的常见类型:,若,a,n,是等差数列,则,=,(,d,0);,=,(,-,);,=,-,.,4.,(2013课标全国理改编,7,5分,0.718)设等差数列,a,n,的前,n,项和为,S,n,若,S,m,-1,=-2,S,m,=0,S,m,+1,=3,则,m,=,.,答案,5,解析,S,m,-1,=-2,S,m,=0,S,m,+1,=3,a,m,=,S,m,-,S,m,-1,=2,a,m,+1,=,S,m,+1,-,S,m,=3,公差,d,=,a,m,+1,-,a,m,=1,由,S,n,=,na,1,+,d,=,na,1,+,得,由得,a,1,=,代入可得,m,=5.,5.,(2015重庆改编,2,5分)在等差数列,a,n,中,若,a,2,=4,a,4,=2,则,a,6,=,.,答案,0,解析,设数列,a,n,的公差为,d,由,a,4,=,a,2,+2,d,a,2,=4,a,4,=2,得2=4+2,d,d,=-1,a,6,=,a,4,+2,d,=0.,6.,(2014福建改编,3,5分)等差数列,a,n,的前,n,项和为,S,n,若,a,1,=2,S,3,=12,则,a,6,等于,.,答案,12,解析,S,3,=,=3,a,2,=12,a,2,=4.,a,1,=2,d,=,a,2,-,a,1,=4-2=2.,a,6,=,a,1,+5,d,=12.,7.,(2017课标全国文,17,12分)记,S,n,为等比数列,a,n,的前,n,项和.已知,S,2,=2,S,3,=-6.,(1)求,a,n,的通项公式;,(2)求,S,n,并判断,S,n,+1,S,n,S,n,+2,是否成等差数列.,e,解析,本题考查等差、等比数列.,(1)设,a,n,的公比为,q,由题设可得,解得,q,=-2,a,1,=-2.,故,a,n,的通项公式为,a,n,=(-2),n,.,(2)由(1)可得,S,n,=,=-,+(-1),n,.,由于,S,n,+2,+,S,n,+1,=-,+(-1),n,=2,=2,S,n,故,S,n,+1,S,n,S,n,+2,成等差数列.,方法总结,等差、等比数列的常用公式:,(1)等差数列:,递推关系式:,a,n,+1,-,a,n,=,d,常用于等差数列的证明.,通项公式:,a,n,=,a,1,+(,n,-1),d,.,前,n,项和公式:,S,n,=,=,na,1,+,d,.,(2)等比数列:,递推关系式:,=,q,(,q,0),常用于等比数列的证明.,通项公式:,a,n,=,a,1,q,n,-1,.,前,n,项和公式:,S,n,=,(3)在证明,a,b,c,成等差、等比数列时,还可以利用等差中项:,=,b,或等比中项:,a,c,=,b,2,来证明.,8.,(2016课标全国,17,12分)等差数列,a,n,中,a,3,+,a,4,=4,a,5,+,a,7,=6.,(1)求,a,n,的通项公式;,(2)设,b,n,=,a,n,求数列,b,n,的前10项和,其中,x,表示不超过,x,的最大整数,如0.9=0,2.6=2.,解析,(1)设数列,a,n,的公差为,d,由题意有2,a,1,+5,d,=4,a,1,+5,d,=3.,解得,a,1,=1,d,=,.,(3分),所以,a,n,的通项公式为,a,n,=,.,(5分),(2)由(1)知,b,n,=,.,(6分),当,n,=1,2,3时,1,2,b,n,=1;,当,n,=4,5时,2,3,b,n,=2;,当,n,=6,7,8时,3,4,b,n,=3;,当,n,=9,10时,4,0,a,7,+,a,10,0,即,a,8,0.,又,a,8,+,a,9,=,a,7,+,a,10,0,a,9,M,;或者存在正整数,m,使得,c,m,c,m,+1,c,m,+2,是等差数列.,解析,本题考查等差数列,不等式,合情推理等知识,考查综合分析,归纳抽象,推理论证能力.,(1),c,1,=,b,1,-,a,1,=1-1=0,c,2,=max,b,1,-2,a,1,b,2,-2,a,2,=max1-2,1,3-2,2=-1,c,3,=max,b,1,-3,a,1,b,2,-3,a,2,b,3,-3,a,3,=max1-3,1,3-3,2,5-3,3=-2.,当,n,3时,(,b,k,+1,-,na,k,+1,)-(,b,k,-,na,k,)=(,b,k,+1,-,b,k,)-,n,(,a,k,+1,-,a,k,)=2-,n,0时,取正整数,m,则当,n,m,时,nd,1,d,2,因此,c,n,=,b,1,-,a,1,n,.,此时,c,m,c,m,+1,c,m,+2,是等差数列.,当,d,1,=0时,对任意,n,1,c,n,=,b,1,-,a,1,n,+(,n,-1)max,d,2,0=,b,1,-,a,1,+(,n,-1)(max,d,2,0-,a,1,).,此时,c,1,c,2,c,3,c,n,是等差数列.,当,d,1,时,有,nd,1,max,故当,n,m,时,M,.,解后反思,解决数列的相关题时,可通过对某些项的观察,分析和比较,发现它们的相同性质或,变化规律,再利用综合法进行推理论证.,4.,(2016课标全国,17,12分)已知,a,n,是公差为3的等差数列,数列,b,n,满足,b,1,=1,b,2,=,a,n,b,n,+1,+,b,n,+1,=,nb,n,.,(1)求,a,n,的通项公式;,(2)求,b,n,的前,n,项和.,解析,(1)由已知,a,1,b,2,+,b,2,=,b,1,b,1,=1,b,2,=,得,a,1,=2,(3分),所以数列,a,n,是首项为2,公差为3的等差数列,通项公式为,a,n,=3,n,-1.,(5分),(2)由(1)和,a,n,b,n,+1,+,b,n,+1,=,nb,n,得,b,n,+1,=,(7分),因此,b,n,是首项为1,公比为,的等比数列.,(9分),记,b,n,的前,n,项和为,S,n,则,S,n,=,=,-,.,(12分),评析,本题主要考查了等差数列及等比数列的定义,能准确写出,a,n,的表达式是关键.,5.,(2016天津理,18,13分)已知,a,n,是各项均为正数的等差数列,公差为,d,.对任意的,n,N,*,b,n,是,a,n,和,a,n,+1,的等比中项.,(1)设,c,n,=,-,n,N,*,求证:数列,c,n,是等差数列;,(2)设,a,1,=,d,T,n,=,(-1),k,n,N,*,求证:,.,证明,(1)由题意得,=,a,n,a,n,+1,有,c,n,=,-,=,a,n,+1,a,n,+2,-,a,n,a,n,+1,=2,da,n,+1,因此,c,n,+1,-,c,n,=2,d,(,a,n,+2,-,a,n,+1,)=2,d,2,所以,c,n,是等差数列.,(2),T,n,=(-,+,)+(-,+,)+,+(-,+,),=2,d,(,a,2,+,a,4,+,+,a,2,n,),=2,d,=2,d,2,n,(,n,+1).,所以,=,=,=,0的等差数列,a,n,的四个命题:,p,1,:数列,a,n,是递增数列;,p,2,:数列,na,n,是递增数列;,p,3,:数列,是递增数列;,p,4,:数列,a,n,+3,nd,是递增数列.,其中的真命题为,.,C,组 教师专用题组,答案,p,1,p,4,解析,a,n,是等差数列,则,a,n,=,a,1,+(,n,-1),d,=,dn,+,a,1,-,d,因为,d,0,所以,a,n,是递增数列,故,p,1,是真命题;对,p,2,举反例,令,a,1,=-3,a,2,=-2,d,=1,则,a,1,2,a,2,故,na,n,不是递增数列,p,2,不是真命题;,=,d,+,当,a,1,-,d,0,时,递减,p,3,不是真命题;,a,n,+3,nd,=4,nd,+,a,1,-,d,4,d,0,a,n,+3,nd,是递增数列,p,4,是真命题.故,p,1,p,4,是真,命题.,2.,(2016课标全国理,17,12分),S,n,为等差数列,a,n,的前,n,项和,且,a,1,=1,S,7,=28.记,b,n,=lg,a,n,其中,x,表,示不超过,x,的最大整数,如0.9=0,lg 99=1.,(1)求,b,1,b,11,b,101,;,(2)求数列,b,n,的前1 000项和.,解析,(1)设,a,n,的公差为,d,据已知有7+21,d,=28,解得,d,=1.,所以,a,n,的通项公式为,a,n,=,n,.,b,1,=lg 1=0,b,11,=lg 11=1,b,101,=lg 101=2.,(6分),(2)因为,b,n,=,(9分),所以数列,b,n,的前1 000项和为1,90+2,900+3,1=1 893.,(12分),疑难突破,充分理解,x,的意义,求出,b,n,的表达式,从而求出,b,n,的前1 000项和.,评析,本题主要考查了数列的综合运用,同时对考生创新能力进行了考查,充分理解,x,的意义,是解题关键.,3.,(2014四川,19,12分)设等差数列,a,n,的公差为,d,点(,a,n,b,n,)在函数,f,(,x,)=2,x,的图象上(,n,N,*,).,(1)若,a,1,=-2,点(,a,8,4,b,7,)在函数,f,(,x,)的图象上,求数列,a,n,的前,n,项和,S,n,;,(2)若,a,1,=1,函数,f,(,x,)的图象在点(,a,2,b,2,)处的切线在,x,轴上的截距为2-,求数列,的前,n,项和,T,n,.,解析,(1)由已知得,b,7,=,b,8,=,=4,b,7,有,=4,=,.,解得,d,=,a,8,-,a,7,=2.,所以,S,n,=,na,1,+,d,=-2,n,+,n,(,n,-1)=,n,2,-3,n,.,(2)函数,f,(,x,)=2,x,在(,a,2,b,2,)处的切线方程为,y,-,=(,ln 2)(,x,-,a,2,),它在,x,轴上的截距为,a,2,-,.,由题意得,a,2,-,=2-,解得,a,2,=2.,所以,d,=,a,2,-,a,1,=1.,从而,a,n,=,n,b,n,=2,n,.,所以,T,n,=,+,+,+,+,+,2,T,n,=,+,+,+,+,.,因此,2,T,n,-,T,n,=1+,+,+,+,-,=2-,-,=,.,所以,T,n,=,.,评析,本题考查等差数列与等比数列的概念、等差数列与等比数列通项公式与前,n,项和、导数,的几何意义等基础知识,考查运算求解能力.,一、填空题(每题5分,共35分),1.,(2017江苏苏州期末,8)设,S,n,是等差数列,a,n,的前,n,项和,若,a,2,=7,S,7,=-7,则,a,7,的值为,.,三年模拟,A组 2015,2017年高考模拟,基础题组,(时间：40分钟 分值：60分),答案,-13,解析,设等差数列,a,n,的公差为,d,由已知可得,解得,所以,a,7,=,a,1,+6,d,=-13.,2.,(2017江苏南京、盐城一模,8)设,a,n,是等差数列,若,a,4,+,a,5,+,a,6,=21,则,S,9,=,.,答案,63,解析,a,n,是等差数列,a,4,+,a,5,+,a,6,=21,a,4,+,a,5,+,a,6,=3,a,5,=21,解得,a,5,=7,S,9,=,(,a,1,+,a,9,)=9,a,5,=63.,3.,(2017江苏淮阴中学第一学期期中,6)已知,S,n,是等差数列,a,n,的前,n,项和,且,S,11,=35+,S,6,则,S,17,的值为,.,答案,119,解析,由,S,11,=35+,S,6,得,S,11,-,S,6,=35,故,a,11,+,a,10,+,a,9,+,a,8,+,a,7,=35,从而5,a,9,=35,所以,a,9,=7,所以,S,17,=,17=,a,9,17=7,17=119.,4.,(2016江苏南京、盐城二模,6)设公差不为0的等差数列,a,n,的前,n,项和为,S,n,.若,S,3,=,且,S,1,S,2,S,4,成,等比数列,则,a,10,等于,.,答案,19,解析,设公差为,d,由题设可得,解得,a,1,=1,d,=2,从而,a,10,=,a,1,+9,d,=1+9,2=19.,5.,(2016江苏扬州中学四模,9)各项均为实数的等差数列的公差为2,其首项的平方与其余各项之,和不超过33,则这样的数列至多有,项.,答案,7,解析,记这个数列为,a,n,则由题意可得,+,a,2,+,a,3,+,+,a,n,=,+,=,+(,n,-1)(,a,1,+,n,)=,+(,n,-1),a,1,+,n,(,n,-1)=,+,n,(,n,-1)-,=,+,33,为了使,n,尽量大,故,=0,33,(,n,-1)(3,n,+1),132,当,n,=6时,5,190,数列,a,n,满足,a,n,+1,=|,p,-,a,n,|+2,a,n,+,p,n,N,*,.,(1)若,a,1,=-1,p,=1.,求,a,4,的值;,求数列,a,n,的前,n,项和,S,n,;,(2)若数列,a,n,中存在三项,a,r,a,s,a,t,(,r,s,t,N,*,r,s,0,所以,a,n,+1,a,n,即,a,n,为单调递增数列.,(i)当,1时,有,a,1,p,于是,a,n,a,1,p,所以,a,n,+1,=|,p,-,a,n,|+2,a,n,+,p,=,a,n,-,p,+2,a,n,+,p,=3,a,n,所以,a,n,=3,n,-1,a,1,.,若,a,n,中存在三项,a,r,a,s,a,t,(,r,s,t,N,*,r,s,t,)依次成等差数列,则有2,a,s,=,a,r,+,a,t,即2,3,s,-1,=3,r,-1,+3,t,-1,.(*),因为,s,t,-1,所以2,3,s,-1,=,3,s,3,t,-1,3,r,-1,+3,t,-1,即(*)式不成立.,故此时数列,a,n,中不存在三项依次成等差数列.,(ii)当-1,1时,有-,p,a,1,p,于是当,n,2时,a,n,a,2,p,从而,a,n,+1,=|,p,-,a,n,|+2,a,n,+,p,=,a,n,-,p,+2,a,n,+,p,=3,a,n,.,所以,a,n,=3,n,-2,a,2,=3,n,-2,(,a,1,+2,p,)(,n,2).,若,a,n,中存在三项,a,r,a,s,a,t,(,r,s,t,N,*,r,s,t,)依次成等差数列,同(i)可知,r,=1,于是有2,3,s,-2,(,a,1,+2,p,)=,a,1,+3,t,-2,(,a,1,+2,p,).,因为2,s,t,-1,所以,=2,3,s,-2,-3,t,-2,=,3,s,-,3,t,-1,0.,因为2,3,s,-2,-3,t,-2,是整数,所以,-1,于是,a,1,-,a,1,-2,p,即,a,1,-,p,与-,p,a,1,p,相矛盾.,故此时数列,a,n,中不存在三项依次成等差数列.,(iii)当,-1时,则有,a,1,-,p,p,a,1,+,p,0,于是,a,2,=|,p,-,a,1,|+2,a,1,+,p,=,p,-,a,1,+2,a,1,+,p,=,a,1,+2,p,a,3,=|,p,-,a,2,|+2,a,2,+,p,=|,p,+,a,1,|+2,a,1,+5,p,=-,p,-,a,1,+2,a,1,+5,p,=,a,1,+4,p,此时有,a,1,a,2,a,3,成等差数列.,综上可知:,-1.,9.,(2016江苏扬州中学质检,20)已知数列,a,n,满足,a,1,=,x,a,2,=3,x,S,n,+1,+,S,n,+,S,n,-1,=3,n,2,+2(,n,2,n,N,*,),S,n,是数,列,a,n,的前,n,项和.,(1)若数列,a,n,为等差数列.,(i)求数列的通项,a,n,;,(ii)若数列,b,n,满足,b,n,=,数列,c,n,满足,c,n,=,t,2,b,n,+2,-,tb,n,+1,-,b,n,试比较数列,b,n,的前,n,项和,B,n,与,c,n,的前,n,项和,C,n,的大小;,(2)若对任意,n,N,*,a,n,0,其前,n,项和,B,n,0,又,c,n,=,t,2,b,n,+2,-,tb,n,+1,-,b,n,=(16,t,2,-4,t,-1),b,n,所以其前,n,项和,C,n,=(16,t,2,-4,t,-1),B,n,所以,C,n,-,B,n,=2(8,t,2,-2,t,-1),B,n,当,t,时,C,n,B,n,;,当,t,=-,或,t,=,时,C,n,=,B,n,;,当-,t,时,C,n,B,n,.,(2)由,S,n,+1,+,S,n,+,S,n,-1,=3,n,2,+2(,n,2,n,N,*,)知,S,n,+2,+,S,n,+1,+,S,n,=3(,n,+1),2,+2(,n,N,*,),两式作差,得,a,n,+2,+,a,n,+1,+,a,n,=6,n,+3(,n,2,n,N,*,),所以,a,n,+3,+,a,n,+2,+,a,n,+1,=6(,n,+1)+3(,n,N,*,),再作差得,a,n,+3,-,a,n,=6(,n,2,n,N,*,),所以,当,n,=1,k,N,*,时,a,n,=,a,1,=,x,;,当,n,=3,k,-1,k,N,*,时,a,n,=,a,3,k,-1,=,a,2,+(,k,-1),6=3,x,+6,k,-6=2,n,+3,x,-4;,当,n,=3,k,k,N,*,时,a,n,=,a,3,k,=,a,3,+(,k,-1),6=14-9,x,+6,k,-6=2,n,-9,x,+8;,当,n,=3,k,+1,k,N,*,时,a,n,=,a,3,k,+1,=,a,4,+(,k,-1),6=1+6,x,+6,k,-6=2,n,+6,x,-7.,因为对任意的,n,N,*,a,n,a,n,+1,恒成立,所以,a,1,a,2,且,a,3,k,-1,a,3,k,a,3,k,+1,a,3,k,+2,所以,解得,x,0,a,2,+,b,2,0,则,a,3,+,b,3,的取值范围是,.,答案,(-,-2),解析,由题意知,a,2,+,b,2,=,a,1,+2+2,b,1,0,所以0,a,1,+,b,1,-2-,b,1,b,1,-2,b,2,=2,b,1,-4,a,3,+,b,3,=,a,2,+,2+2,b,2,=,a,2,+,b,2,+2+,b,2,0,当,n,2,n,N,*,时,f,(,n,)是递增数列,f,(,n,)的最小值是,f,(2)=,.,一、填空题,1.,(2017江苏六校联考,11)已知函数,f,(,x,)=,x,3,+,x,等差数列,a,n,满足:,f,(,a,2,-1)=2,f,(,a,2 016,-3)=-2,S,n,是其前,n,项和,则,S,2 017,=,.,C,组 2015,2017年高考模拟,创新题组,答案,4 034,解析,由函数,f,(,x,)=,x,3,+,x,可得,f,(,x,)为奇函数,且单调递增,由,f,(,a,2,-1)=2,f,(,a,2 016,-3)=-2,可得,f,(,a,2,-1)=,f,-(,a,2 016,-3),即,a,2,-1=-(,a,2 016,-3),所以,a,2,+,a,2 016,=4,从而,S,2 017,=,2 017=,2 017=2,2 017=4 034.,方法点拨,当数列,a,n,为等差数列时,若,m,+,n,=,p,+,q,=2,k,(,m,n,p,q,k,N,*,),则,a,m,+,a,n,=,a,p,+,a,q,=2,a,k,.,2.,(2016江苏启东中学阶段测试)已知数列,a,n,满足,a,1,=1,且,a,n,+1,=,a,n,+,n,N,*,则,k,(,a,2 015,-,a,k,)=,.,答案,解析,由已知得当,n,2时,a,n,=,a,n,-1,+,=,a,n,-2,+,+,=,=1+,+,+,当,n,=1时,a,1,=1,a,n,=1+,+,+,(,n,N,*,).,k,(,a,2 015,-,a,k,)=,k,k,(,a,2 015,-,a,k,),=,+2,+,+2 014,=,+(1+2),+(1+2+3),+,+(1+2+,+,k,),+,+(1+2+,+2 014),=,+,+,+,+,+,+,=,.,二、解答题,3.,(2016江苏南京调研)已知数列,a,n,b,n,满足:,b,n,=,a,n,+1,-,a,n,(,n,N,*,).,(1)若,a,1,=1,b,n,=,n,求数列,a,n,的通项公式;,(2)若,b,n,+1,b,n,-1,=,b,n,(,n,2),且,b,1,=1,b,2,=2.,记,c,n,=,a,6,n,-1,(,n,1),求证:数列,c,n,为等差数列;,若数列,中任意一项的值均未在该数列中重复出现无数次,求首项,a,1,应满足的条件.,解析,(1)当,n,2时,a,n,=,a,1,+(,a,2,-,a,1,)+(,a,3,-,a,2,)+,+(,a,n,-,a,n,-1,)=,a,1,+,b,1,+,b,2,+,+,b,n,-1,=,-,+1,因为,a,1,=1也满足上式,所以数列,a,n,的通项公式是,a,n,=,-,+1.,(2)证明:因为对任意的,n,N,*,有,b,n,+6,=,=,=,=,b,n,所以,c,n,+1,-,c,n,=,a,6,n,+5,-,a,6,n,-1,=,b,6,n,-1,+,b,6,n,+,b,6,n,+1,+,b,6,n,+2,+,b,6,n,+3,+,b,6,n,+4,=,b,1,+,b,2,+,b,3,+,b,4,+,b,5,+,b,6,=1+2+2+1+,+,=7,所以,数列,c,n,为等差数列.,设(,d,i,),n,=,a,6(,n,-1)+,i,(,n,N,*,)(其中,i,为常数且,i,1,2,3,4,5,6),所以(,d,i,),n,+1,-(,d,i,),n,=,a,6(,n,-1)+6+,i,-,a,6(,n,-1)+,i,=,b,6(,n,-1)+,i,+,b,6(,n,-1)+,i,+1,+,b,6(,n,-1)+,i,+2,+,b,6(,n,-1)+,i,+3,+,b,6(,n,-1)+,i,+4,+,b,6(,n,-1)+,i,+5,=7,所以数列,a,6(,n,-1)+,i,为以7为公差的等差数列,设,f,i,(,k,)=,=,=,+,其中,k,0,i,为1,2,3,4,5,6中的一个常数,a.当,a,i,=,时,对任意的,n,=6,k,+,i,有,=,.,当,i,=1时,a,1,=,1=,;,当,i,=2时,a,2,=,2=,此时,a,1,=,a,2,-,b,1,=,;,当,i,=3时,a,3,=,3=,此时,a,1,=,a,3,-(,b,1,+,b,2,)=,;,当,i,=4时,a,4,=,4=,此时,a,1,=,a,4,-(,b,1,+,b,2,+,b,3,)=-,;,当,i,=5时,a,5,=,5=,此时,a,1,=,a,5,-(,b,1,+,b,2,+,b,3,+,b,4,)=-,;,当,i,=6时,a,6,=,6=7,此时,a,1,=,a,6,-(,b,1,+,b,2,+,b,3,+,b,4,+,b,5,)=-,.,b.当,a,i,时,f,i,(,k,+1)-,f,i,(,k,)=,-,=,.,若,a,i,则对任意的,k,N有,f,i,(,k,+1),f,i,(,k,),所以数列,为递减数列;,若,a,i,f,i,(,k,),所以数列,为递增数列,记集合,B,=,.当,a,1,B,时,数列,中必有某数重复出现无数次,不符合题意;当,a,1,B,时,数列,(,i,=1,2,3,4,5,6)均为单调数列,任意一个数在这6个数列中最多出现一次,即任,意一个数在数列,中最多出现6次,所以若数列,中任意一项的值均未在该数列中重复出,现无数次,则首项,a,1,R,B,.,