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2018-11-8,#,栏目索引,高考导航,核心题型突破,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,*,微专题,13,数列中的探索性问题,微专题13数列中的探索性问题,题型一新定义数列的探究型问题,例1,(2018江苏扬州高三模拟) 已知数列,a,n,中,a,1,=1,前,n,项和为,S,n,若对任意,的,n,N,*,均有,S,n,=,a,n,+,k,-,k,(,k,是常数,且,k,N,*,)成立,则称数列,a,n,为“,H,(,k,)数列”.,(1)若数列,a,n,为“,H,(1)数列”,求数列,a,n,的前,n,项和,S,n,;,(2)若数列,a,n,为“,H,(2)数列”,且,a,2,为整数,试问:是否存在数列,a,n,使得|,-,a,n,-,1,a,n,+1,|,40对任意,n,2,n,N,*,成立?如果存在,求出这样数列,a,n,的,a,2,的所有可能,值,如果不存在,请说明理由.,解析,(1)因为数列,a,n,为“,H,(1)数列”,所以,S,n,=,a,n,+1,-1,故,S,n,-1,=,a,n,-1(,n,2),两式,相减得,a,n,+1,=2,a,n,(,n,2),在,S,n,=,a,n,+1,-1中令,n,=1,则可得,a,2,=2,故,a,2,=2,a,1,.,所以,=2(,n,N,*,),所以数列,a,n,为等比数列,所以,a,n,=2,n,-1,所以,S,n,=2,n,-1.,(2)由题意得,S,n,=,a,n,+2,-2,故,S,n,-1,=,a,n,+1,-2(,n,2),两式相减得,a,n,+2,=,a,n,+1,+,a,n,(,n,2),所以,当,n,2时,-,a,n,a,n,+2,=,-,a,n,(,a,n,+1,+,a,n,)=,a,n,+1,(,a,n,+1,-,a,n,)-,又因为,a,n,+1,-,a,n,=,a,n,-1,(,n,3),所以,-,a,n,a,n,+2,=,a,n,+1,(,a,n,+1,-,a,n,)-,=,a,n,+1,a,n,-1,-,所以|,-,a,n,a,n,+2,|=|,-,a,n,+1,a,n,-1,|(,n,3),所以当,n,3时,数列|,-,a,n,+1,a,n,-1,|是常数列,所以|,-,a,n,+1,a,n,-1,|=|,-,a,2,a,4,|(,n,3),因为,a,4,=,a,3,+,a,2,所以|,-,a,n,+1,a,n,-1,|=|,-,a,2,a,3,-,|(,n,3).,在,S,n,=,a,n,+2,-2中令,n,=1,则可得,a,3,=3,所以|9-3,a,2,-,|,40,又,n,=2时|,-,a,1,a,3,|=|,-3|,40,且,a,2,为整数,所以可解得,a,2,=0,1,2,3,4,5,-6.,【方法归纳】对于新定义数列中的探究性问题,读懂、理解新数列的定义,是重点.一般而言,这类题目考查的难点已在新定义中体现,后续反而不会太,难,但需要具备举一反三的能力,结合原有数列知识去探求出题目所要求的条,件,大胆尝试、总结.,1-1,(2018泰州中学高三检测)数列,a,n,对于确定的正整数,m,若存在正整数,n,使得,a,m,+,n,=,a,m,+,a,n,成立,则称数列,a,n,为“,m,阶可分拆数列”.,(1)设,a,n,是首项为2,公差为2的等差数列,证明,a,n,为“3阶可分拆数列”;,(2)设数列,a,n,的前,n,项和为,S,n,=2,n,-,a,(,a,0),若数列,a,n,为“1阶可分拆数列”,求,实数,a,的值;,(3)设,a,n,=2,n,+,n,2,+12,试探求是否存在,m,使得若数列,a,n,为“,m,阶可分拆数列”.,若存在,请求出所有,m,;若不存在,请说明理由.,解析,(1)证明:,a,n,=2+2(,n,-1)=2,n,a,3,=6,则,a,3+,n,=2,(3+,n,)=6+2,n,=,a,3,+,a,n,.,a,n,为“3阶可分拆数列”.,(2),S,n,=2,n,-,a,(,a,0),a,1,=,S,1,=2-,a,n,2时,a,n,=,S,n,-,S,n,-1,=2,n,-,a,-(2,n,-1,-,a,)=2,n,-1,.,数列,a,n,为“1阶可分拆数列”,a,n,+1,=,a,1,+,a,n,2,n,=2-,a,+2,n,-1,a,=2-2,n,-1,.,令,n,=1时,a,=1.,(3)假设数列,a,n,为“,m,阶可分拆数列”.,则,a,m,+,n,=,a,m,+,a,n,成立,2,n,+,m,+(,n,+,m,),2,+12=2,m,+,m,2,+12+2,n,+,n,2,+12,化为2,n,+,m,+2,mn,=2,m,+2,n,+12,(2,m,-1)(2,n,-1)+2,mn,=13.,可得:,m,=1,n,=3;,m,=2,n,不存在;,m,=3,n,=1;,m,4时,n,不存在.,只有两组:,m,=1,n,=3;,m,=3,n,=1.,题型二探究数列中是否存在满足条件的项的问题,例2,(2018扬州高三考前调研)已知无穷数列,a,n,的各项都不为零,其前,n,项,和为,S,n,且满足,a,n,a,n,+1,=,S,n,(,n,N,*,),数列,b,n,满足,b,n,=,其中,t,为正整数.,(1)求,a,2 018,;,(2)若不等式,+,S,n,+,S,n,+1,对任意,n,N,*,都成立,求首项,a,1,的取值范围;,(3)若首项,a,1,是正整数,则数列,b,n,中的任意一项是否总可以表示为数列,b,n,中,的其他两项之积?若是,请给出一种表示方式;若不是,请说明理由.,解析,(1)令,n,=1,则,a,1,a,2,=,S,1,即,a,1,a,2,=,a,1,又,a,1,0,故,a,2,=1.,由,a,n,a,n,+1,=,S,n,得,a,n,+1,a,n,+2,=,S,n,+1,两式相减得(,a,n,+2,-,a,n,),a,n,+1,=,a,n,+1,又,a,n,+1,0,故,a,n,+2,-,a,n,=1.,所以数列,a,2,n,是首项为1、公差为1的等差数列.,所以,a,2018,=,a,2,+,1=1 009.,(2)由(1)知,数列,a,2,n,是首项为1,公差为1的等差数列;数列,a,2,n,-1,是首项为,a,1,公,差为1的等差数列,故,a,n,=,所以,S,n,=,当,n,是奇数时,+,S,n,+,S,n,+1,即,+,+,即,-2,a,1,对任意正奇数,n,恒成立,所以,-2,a,1,0,即0,a,1,2;,当,n,是偶数时,+,S,n,+,S,n,+1,即,+,+,即,-,a,1,对任意正偶数,n,恒成立,所以,-,a,1,1,即,a,1,.,综合得:0,a,1,.,(3)由数列,a,2,n,是首项为1、公差为1的等差数列,数列,a,2,n,-1,是首项为正整数,a,1,、公差为1的等差数列知,数列,a,n,的各项都是正整数,设,b,n,=,b,m,b,k,即,=,即,a,m,=,取,k,=,n,+2,则,a,k,-,a,n,=1,故,a,m,=,a,n,(,a,n,+2,+,t,),不妨设,m,是偶数,则,=,a,n,(,a,n,+2,+,t,)一定是整数,故当,n,是偶数时,方程,b,n,=,b,m,b,k,的一组解是,故当,n,是奇数时,方程,b,n,=,b,m,b,k,的一组解是,所以,数列,b,n,中的任意一项总可以表示为数列,b,n,中的其他两项之积.,【方法归纳】(1)此类问题常与函数、方程、不等式等知识相互关联渗透,解题时注意方程思想、整体思想、分类讨论思想以及数形结合思想等的灵,活运用.,(2)数列中的不等式恒成立问题与函数问题中的不等式恒成立问题的解法一,致,即都是利用分离参数转化为求最值,但要注意数列的单调性与函数单调性,的研究是有区别的,关键是数列的定义域是正整数集(或其子集).,2-1,(2018江苏盐城高三期中)已知数列,a,n,满足,a,1,=-1,a,2,=1,且,a,n,+2,=,a,n,(,n,N,*,).,(1)求,a,5,+,a,6,的值;,(2),S,n,为数列,a,n,的前,n,项的和,求,S,n,;,(3)设,b,n,=,a,2,n,-1,+,a,2,n,是否存正整数,i,j,k,(,i,j,j,且,k,j,Z,k,j,+1.,当,k,j,+2时,b,k,b,j,+2,若,b,i,b,j,b,k,成等差数列,则,b,i,=2,b,j,-,b,k,2,b,j,-,b,j,+2,=2,-,=-,-,0,此与,b,n,0矛盾.故此时不存在这样的等差数列;,当,k,=,j,+1时,b,k,=,b,j,+1,若,b,i,b,j,b,k,成等差数列,则,b,i,=2,b,j,-,b,k,=2,b,j,-,b,j,+1,=2,-,=,-,又,i,2总成立?若存在,求,出,q,的范围;若不存在,请说明理由.,解析,(1)因为,a,1,=1,a,3,+,a,5,=20,所以,q,4,+,q,2,-20=0,所以,q,2,=4或,q,2,=-5(舍去).,所以,=1+,q,4,=17.,(2)若,a,2,a,1,a,3,或,a,3,a,1,a,2,成等差数列,则2,a,1,=,a,3,+,a,2,q,2,+,q,-2=0,解得,q,=-2或1(舍去);,若,a,1,a,3,a,2,或,a,2,a,3,a,1,成等差数列,则2,a,3,=,a,1,+,a,2,2,q,2,-,q,-1=0,解得,q,=-,或1(舍去);,若,a,3,a,2,a,1,成等差数列,则2,a,2,=,a,3,+,a,1,q,2,-2,q,+1=0,解得,q,=1(舍去).,综上,q,=-2或-,.,(3)由,-20(,c,0),可得,0,故等价于,c,S,n,0,所以,S,n,1,得到,c,1时,S,2,22,c,不可能成立;,当,q,2,得,q,n,log,q,(2,q,-1).,因为,q,1,即当,n,log,q,(2,q,-1)时,S,n,2,所以,S,n,2,c,不可能成立;,当,q,=,时,由,2,c,即1-,1-,c,则当,n,lo,(1-,c,)时,S,n,2,c,不成立;,当0,q,时,S,n,=,所以当,c,1时,c,S,n,2总成立,q,的取值范围,为,.,【方法归纳】此类问题往往结合整数方程、函数的性质等进行求解,需要,在理解题意的基础上不断地尝试探索,往往是从特殊到一般寻找规律,会使用,到列举法、特殊值法、代入验证等方法.总之,我们必须仔细审题,合情推理,恰当转化,透过现象看本质.,3-1,(2018江苏海安高级中学高三月考)已知数列,a,n,中,首项,a,1,=1,a,2,=,a,a,n,+1,=,k,(,a,n,+,a,n,+2,)对任意正整数,n,都成立,数列,a,n,的前,n,项和为,S,n,.,(1)若,k,=,且,S,18,=171,求实数,a,的值;,(2)是否存在实数,k,使数列,a,n,是公比不为1的等比数列,且任意相邻三项,a,n,a,n,+,1,a,n,+2,按某顺序排列后成等差数列.若存在,求出所有的,k,的值;若不存在,请说明,理由;,(3)若,k,=-,求,S,n,(用,a,n,表示).,解析,(1)当,k,=,时,由,a,n,+1,=,k,(,a,n,+,a,n,+2,)得,a,n,+1,=,(,a,n,+,a,n,+2,),即,a,n,+2,-,a,n,+1,=,a,n,+1,-,a,n,所以数列,a,n,为等差数列,公差为,d,=,a,2,-,a,1,=,a,-1,数列,a,n,的前,n,项和为,S,n,=,n,+,(,a,-1),由,S,18,=171得171=18+,(,a,-1),解得,a,=2.,(2)设存在,k,使数列,a,n,为等比数列,则其公比为,q,=,=,a,a,n,=,a,n,-1,a,n,+1,=,a,n,a,n,+2,=,a,n,+1,.,a,n,+1,为等差中项,则2,a,n,+1,=,a,n,+,a,n,+2,即2,a,n,=,a,n,-1,+,a,n,+1,解得,a,=1,与已知不符,舍去;,若,a,n,为等差中项,则2,a,n,=,a,n,+1,+,a,n,+2,即2,a,n,-1,=,a,n,+,a,n,+1,即,a,2,+,a,-2=0,解得,a,=-2或,a,=1,(舍),此时由,a,n,+1,=,k,(,a,n,+,a,n,+2,)得,a,n,=,k,(,a,n,-1,+,a,n,+1,),即,a,=,k,(1+,a,2,),故,k,=,=-,;,若,a,n,+2,为等差中项,则2,a,n,+2,=,a,n,+,a,n,+1,即2,a,n,+1,=,a,n,-1,+,a,n,即2,a,2,-,a,-1=0,解得,a,=-,或,a,=,1(舍),仿得,k,=,=-,.,综上,满足要求的实数,k,有且仅有一个,k,=-,.,(3)当,k,=-,时,a,n,+1,=-,(,a,n,+,a,n,+2,),所以,a,n,+2,+,a,n,+1,=-(,a,n,+1,+,a,n,),于是,a,n,+3,+,a,n,+2,=-(,a,n,+2,+,a,n,+1,)=,a,n,+1,+,a,n,.,当,n,为偶数时,S,n,=(,a,1,+,a,2,)+(,a,3,+,a,4,)+(,a,5,+,a,6,)+,+(,a,n,-1,+,a,n,)=,(,a,1,+,a,2,)=,;,当,n,为奇数时,S,n,=,a,1,+(,a,2,+,a,3,)+(,a,4,+,a,5,)+,+(,a,n,-1,-,a,n,)=,a,1,+,(,a,2,+,a,3,)=,a,1,+,.-(,a,1,+,a,2,)=1-,(,a,+1)(,n,2),当,n,=1时,也适合该式.,所以,S,n,=,dsfdbsy384y982ythb3oibt4oy39y409705923y09y53b2lkboi2y58wy0ehtoibwoify98wy049ywh4b3oiut89u983yf9ivh98y98sv98hv98ys9f698y9v698yv98x98tb98fyd98gyd98h98ds98nt98d8genklgb4klebtlkb5k 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