江苏专用高考数学一轮复习第二十章推理与证明数学归纳法.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),第二十章推理与证明(数学归纳法),1.,(2014江苏,23,10分,0.20)已知函数,f,0,(,x,)=,(,x,0),设,f,n,(,x,)为,f,n,-1,(,x,)的导数,n,N,*,.,(1)求2,f,1,+,f,2,的值;,(2)证明:对任意的,n,N,*,等式,=,都成立.,A,组 自主命题,江苏卷题组,五年高考,解析,(1)由已知,得,f,1,(,x,)=,f,0,(,x,)=,=,-,于是,f,2,(,x,)=,f,1,(,x,)=,-,=-,-,+,所以,f,1,=-,f,2,=-,+,.,故2,f,1,+,f,2,=-1.,(2)证明:由已知,得,xf,0,(,x,)=sin,x,等式两边分别对,x,求导,得,f,0,(,x,)+,xf,0,(,x,)=cos,x,即,f,0,(,x,)+,xf,1,(,x,)=cos,x,=sin,类似可得,2,f,1,(,x,)+,xf,2,(,x,)=-sin,x,=sin(,x,+),3,f,2,(,x,)+,xf,3,(,x,)=-cos,x,=sin,4,f,3,(,x,)+,xf,4,(,x,)=sin,x,=sin(,x,+2).,下面用数学归纳法证明等式,nf,n,-1,(,x,)+,xf,n,(,x,)=sin,对所有的,n,N,*,都成立.,(i)当,n,=1时,由上可知等式成立.,(ii)假设当,n,=,k,时等式成立,即,kf,k,-1,(,x,)+,xf,k,(,x,)=sin,.,因为,kf,k,-1,(,x,)+,xf,k,(,x,)=,kf,k,-1,(,x,)+,f,k,(,x,)+,xf,k,(,x,)=(,k,+1),f,k,(,x,)+,xf,k,+1,(,x,),=cos,=,sin,所以(,k,+1),f,k,(,x,)+,xf,k,+1,(,x,)=sin,.,因此当,n,=,k,+1时,等式也成立.,综合(i),(ii)可知等式,nf,n,-1,(,x,)+,xf,n,(,x,)=sin,对所有的,n,N,*,都成立.,令,x,=,可得,nf,n,-1,+,f,n,=sin,(,n,N,*,).,所以,=,(,n,N,*,).,2.,(2013江苏,23,10分,0.140)设数列,a,n,:1,-2,-2,3,3,3,-4,-4,-4,-4,即当,n,(,k,N,*,)时,a,n,=(-1),k,-1,k,.记,S,n,=,a,1,+,a,2,+,+,a,n,(,n,N,*,).对于,l,N,*,定义集合,P,l,=,n,|,S,n,是,a,n,的整数倍,n,N,*,且1,n,l,.,(1)求集合,P,11,中元素的个数;,(2)求集合,P,2 000,中元素的个数.,解析,(1)由数列,a,n,的定义得,a,1,=1,a,2,=-2,a,3,=-2,a,4,=3,a,5,=3,a,6,=3,a,7,=-4,a,8,=-4,a,9,=-4,a,10,=-4,a,11,=5,所以,S,1,=1,S,2,=-1,S,3,=-3,S,4,=0,S,5,=3,S,6,=6,S,7,=2,S,8,=-2,S,9,=-6,S,10,=-10,S,11,=-5,从而,S,1,=,a,1,S,4,=0,a,4,S,5,=,a,5,S,6,=2,a,6,S,11,=-,a,11,所以集合,P,11,中元素的个数为5.,(2)先证:,S,i,(2,i,+1),=-,i,(2,i,+1)(,i,N,*,).,事实上,当,i,=1时,S,i,(2,i,+1),=,S,3,=-3,-,i,(2,i,+1)=-3,故原等式成立;,假设,i,=,m,时成立,即,S,m,(2,m,+1),=-,m,(2,m,+1),则,i,=,m,+1时,S,(,m,+1)(2,m,+3),=,S,m,(2,m,+1),+(2,m,+1),2,-(2,m,+2),2,=-,m,(2,m,+1)-4,m,-,3=-(2,m,2,+5,m,+3)=-(,m,+1)(2,m,+3).,综合可得,S,i,(2,i,+1),=-,i,(2,i,+1).于是,S,(,i,+1)(2,i,+1),=,S,i,(2,i,+1),+(2,i,+1),2,=-,i,(2,i,+1)+(2,i,+1),2,=(2,i,+1)(,i,+1).,由上可知,S,i,(2,i,+1),是2,i,+1的倍数,而,a,i,(2,i,+1)+,j,=2,i,+1(,j,=1,2,2,i,+1),所以,S,i,(2,i,+1)+,j,=,S,i,(2,i,+1),+,j,(2,i,+1)是,a,i,(2,i,+1)+,j,(,j,=1,2,2,i,+1)的倍数.又,S,(,i,+1)(2,i,+1),=(,i,+1)(2,i,+1)不是2,i,+2的倍数,而,a,(,i,+1)(2,i,+1)+,j,=-(2,i,+2)(,j,=1,2,2,i,+2),所以,S,(,i,+1)(2,i,+1)+,j,=,S,(,i,+1)(2,i,+1),-,j,(2,i,+2)=(2,i,+1)(,i,+1)-,j,(2,i,+2)不是,a,(,i,+1)(2,i,+1)+,j,(,j,=1,2,2,i,+2)的倍数,故当,l,=,i,(2,i,+1)时,集合,P,l,中元素的个数为1+3+,+(2,i,-1)=,i,2,于是,当,l,=,i,(2,i,+1)+,j,(1,j,2,i,+1)时,集合,P,l,中元素的个数为,i,2,+,j,.,又2 000=31,(2,31+1)+47,故集合,P,2 000,中元素的个数为31,2,+47=1 008.,思路分析,(1)取,n,为1,2,3,11时计算出相应的,S,n,进而作出判断;(2)先利用数学归纳法证明一,般性结论,再求,P,2 000,中元素的个数.,考点数学归纳法,1.,(2017浙江,22,15分)已知数列,x,n,满足:,x,1,=1,x,n,=,x,n,+1,+ln(1+,x,n,+1,)(,n,N,*,).,证明:当,n,N,*,时,(1)0,x,n,+1,0.,当,n,=1时,x,1,=10.,假设,n,=,k,时,x,k,0,那么,n,=,k,+1时,若,x,k,+1,0,则00.,因此,x,n,0(,n,N,*,).所以,x,n,=,x,n,+1,+ln(1+,x,n,+1,),x,n,+1,.,因此0,x,n,+1,0(,x,0).,函数,f,(,x,)在0,+,)上单调递增,所以,f,(,x,),f,(0)=0,因此,-2,x,n,+1,+(,x,n,+1,+2)ln(1+,x,n,+1,)=,f,(,x,n,+1,),0,故2,x,n,+1,-,x,n,(,n,N,*,).,(3)因为,x,n,=,x,n,+1,+ln(1+,x,n,+1,),x,n,+1,+,x,n,+1,=2,x,n,+1,所以,x,n,.,由,2,x,n,+1,-,x,n,得,-,2,0,所以,-,2,2,n,-1,=2,n,-2,故,x,n,.综上,x,n,(,n,N,*,).,方法总结,1.证明数列单调性的方法.,差比法:作差,a,n,+1,-,a,n,然后分解因式,判断符号,或构造函数,利用导数求函数的值域,从而判断其,符号.,商比法:作商,判断,与1的大小,同时注意,a,n,的正负.,数学归纳法.,反证法:例如求证:,n,N,*,a,n,+1,0),则有,n,2时,a,n,=,a,1,a,1,q,n,-1,(其中,a,1,0).,放缩为等比数列:利用不等式性质,把非等比数列,a,n,放缩成等比数列,b,n,求和后,再进行适当,放缩.,2.,(2014广东节选,19,14分)设数列,a,n,的前,n,项和为,S,n,满足,S,n,=2,na,n,+1,-3,n,2,-4,n,n,N,*,且,S,3,=15.,(1)求,a,1,a,2,a,3,的值;,(2)求数列,a,n,的通项公式.,解析,(1)依题有,解得,a,1,=3,a,2,=5,a,3,=7.,(2),S,n,=2,na,n,+1,-3,n,2,-4,n,当,n,2时,S,n,-1,=2(,n,-1),a,n,-3(,n,-1),2,-4(,n,-1).,-并整理得,a,n,+1,=,.,由(1)猜想,a,n,=2,n,+1,下面用数学归纳法证明.,当,n,=1时,a,1,=2+1=3,命题成立;,假设当,n,=,k,时,a,k,=2,k,+1命题成立.,则当,n,=,k,+1时,a,k,+1,=,=,=2,k,+3=2(,k,+1)+1,即当,n,=,k,+1时,结论成立.,综上,n,N,*,a,n,=2,n,+,1.,3.,(2014重庆,22,12分)设,a,1,=1,a,n,+1,=,+,b,(,n,N,*,).,(1)若,b,=1,求,a,2,a,3,及数列,a,n,的通项公式;,(2)若,b,=-1,问:是否存在实数,c,使得,a,2,n,c,a,2,n,+1,对所有,n,N,*,成立?证明你的结论.,解析,(1)解法一:,a,2,=2,a,3,=,+1.,再由题设条件知(,a,n,+1,-1),2,=(,a,n,-1),2,+1.,从而(,a,n,-1),2,是首项为0,公差为1的等差数列,故(,a,n,-1),2,=,n,-1,即,a,n,=,+1(,n,N,*,).,解法二:,a,2,=2,a,3,=,+1,可写为,a,1,=,+1,a,2,=,+1,a,3,=,+1.,因此猜想,a,n,=,+1.,下用数学归纳法证明上式:,当,n,=1时结论显然成立.,假设,n,=,k,时结论成立,即,a,k,=,+1,则,a,k,+1,=,+1=,+1=,+1.,这就是说,当,n,=,k,+1时结论成立.,所以,a,n,=,+1(,n,N,*,).,(2)解法一:设,f,(,x,)=-1,则,a,n,+1,=,f,(,a,n,).,令,c,=,f,(,c,),即,c,=,-1,解得,c,=,.,下用数学归纳法证明加强命题,a,2,n,c,a,2,n,+1,1.,当,n,=1时,a,2,=,f,(1)=0,a,3,=,f,(0)=,-1,所以,a,2,a,3,1,结论成立.,假设,n,=,k,时结论成立,即,a,2,k,c,a,2,k,+1,f,(,a,2,k,+1,),f,(1)=,a,2,即1,c,a,2,k,+2,a,2,.,再由,f,(,x,)在(-,1上为减函数得,c,=,f,(,c,),f,(,a,2,k,+2,),f,(,a,2,)=,a,3,1.,故,c,a,2,k,+3,1,因此,a,2(,k,+1),c,a,2(,k,+1)+1,1.,这就是说,当,n,=,k,+1时结论成立.,综上,符合条件的,c,存在,其中一个值为,c,=,.,解法二:设,f,(,x,)=-1,则,a,n,+1,=,f,(,a,n,).,先证:0,a,n,1(,n,N,*,).,当,n,=1时,结论明显成立.,假设,n,=,k,时结论成立,即0,a,k,1.,易知,f,(,x,)在(-,1上为减函数,从而0=,f,(1),f,(,a,k,),f,(0)=,-11.,即0,a,k,+1,1.这就是说,当,n,=,k,+1时结论成立.故成立.,再证:,a,2,n,a,2,n,+1,(,n,N,*,).,当,n,=1时,a,2,=,f,(1)=0,a,3,=,f,(,a,2,)=,f,(0)=,-1,有,a,2,a,3,即,n,=1时成立.,假设,n,=,k,时,结论成立,即,a,2,k,f,(,a,2,k,+1,)=,a,2,k,+2,a,2(,k,+1),=,f,(,a,2,k,+1,),f,(,a,2,k,+2,)=,a,2(,k,+1)+1,.,这就是说,当,n,=,k,+1时成立.所以对一切,n,N,*,成立.,由得,a,2,n,-1,即(,a,2,n,+1),2,-2,a,2,n,+2,因此,a,2,n,f,(,a,2,n,+1,),即,a,2,n,+1,a,2,n,+2,所以,a,2,n,+1,-1,解得,a,2,n,+1,.,综上,由、知存在,c,=,使,a,2,n,c,1).,(1)讨论,f,(,x,)的单调性;,(2)设,a,1,=1,a,n,+1,=ln(,a,n,+1),证明:,a,n,.,B组,统一命题,省(区、市)卷题组,解析,(1),f,(,x,)的定义域为(-1,+,),f,(,x,)=,.,(2分),(i)当1,a,0,f,(,x,)在(-1,a,2,-2,a,)上是增函数;,若,x,(,a,2,-2,a,0),则,f,(,x,)0,f,(,x,)在(0,+,)上是增函数.,(4分),(ii)当,a,=2时,f,(,x,),0,f,(,x,)=0成立当且仅当,x,=0,f,(,x,)在(-1,+,)上是增函数.,(iii)当,a,2时,若,x,(-1,0),则,f,(,x,)0,f,(,x,)在(-1,0)上是增函数;,若,x,(0,a,2,-2,a,),则,f,(,x,)0,f,(,x,)在(,a,2,-2,a,+,)上是增函数.,(6分),(2)由(1)知,当,a,=2时,f,(,x,)在(-1,+,)上是增函数.,当,x,(0,+,)时,f,(,x,),f,(0)=0,即ln(,x,+1),(,x,0).,又由(1)知,当,a,=3时,f,(,x,)在0,3)上是减函数.,当,x,(0,3)时,f,(,x,),f,(0)=0,即ln(,x,+1),(0,x,3).,(9分),下面用数学归纳法证明,a,n,.,(i)当,n,=1时,由已知,a,1,=1,故结论成立;,(ii)设当,n,=,k,时结论成立,即,ln,=,a,k,+1,=ln(,a,k,+1),ln,=,即当,n,=,k,+1时有,1时,对,x,(0,a,-1有,(,x,)0,(,x,)在(0,a,-1上单调递减,(,a,-1)1时,存在,x,0,使,(,x,),n,-ln(,n,+1).,证明如下:,证法一:上述不等式等价于,+,+,+,x,0.,令,x,=,n,N,+,则,ln,.,下面用数学归纳法证明.,当,n,=1时,ln 2,结论成立.,假设当,n,=,k,时结论成立,即,+,+,+,ln(,k,+1).,那么,当,n,=,k,+1时,+,+,+,+,ln(,k,+1)+,ln(,k,+1)+ln,=ln(,k,+2),即结论成立.,由可知,结论对,n,N,+,成立.,证法二:上述不等式等价于,+,+,+,x,0.,令,x,=,n,N,+,则ln,.,故有ln 2-ln 1,ln 3-ln 2,ln(,n,+1)-ln,n,上述各式相加可得ln(,n,+1),+,+,+,.,结论得证.,评析,本题考查了导数、裂项法求和、数学归纳法、不等式等知识;考查推理论证、运算能力,转化与化归意识、综合分析能力;熟练运用数学归纳法证明或利用函数变形、裂项求和是解题,的关键.,1.,(2017江苏如皋高三上学期教学质量调研(三),21)已知数列,a,n,满足,a,1,=1,a,n,+1,=,(,n,为正整,数).,(1)求,a,2,a,3,a,4,的值,并猜想数列,a,n,的通项公式;,(2)用数学归纳法证明(1)的猜想.,三年模拟,A组 2015,2017年高考模拟,基础题组,(时间：45分钟 分值：40分),解答题(共40分),解析,(1),a,n,+1,=,=2-,a,1,=1,a,2,=2-,=,同理,a,3,=,a,4,=,猜想,a,n,=,(,n,N,*,).,(2)证明:当,n,=1时,猜想成立.,假设当,n,=,k,(,k,N,*,)时,猜想成立,即,a,k,=,则当,n,=,k,+1时,有,a,k,+1,=2-,=2-,=,=,所以当,n,=,k,+1时,猜想成立,综合可得,a,n,=,(,n,N,*,).,2.,(2017江苏南通、扬州、泰州高三第三次模拟考试,23)已知函数,f,0,(,x,)=,(,a,0,ac,-,bd,0).设,f,n,(,x,)为,f,n,-1,(,x,)(,n,N,*,)的导函数.,(1)求,f,1,(,x,),f,2,(,x,);,(2)猜想,f,n,(,x,)的表达式,并证明你的结论.,解析,(1),f,1,(,x,)=,f,0,(,x,)=,=,f,2,(,x,)=,f,1,(,x,)=,=,.,(2)猜想,f,n,(,x,)=,n,N,*,.,证明:当,n,=1时,由(1)知结论成立;,假设当,n,=,k,k,N,*,时结论成立,即有,f,k,(,x,)=,.,当,n,=,k,+1时,f,k,+1,(,x,)=,f,k,(,x,)=,=(-1),k,-1,a,k,-1,(,bc,-,ad,),k,!(,ax,+,b,),-(,k,+1),=,所以当,n,=,k,+1时结论成立.,由得,f,n,(,x,)=,n,N,*,.,3.,(2016江苏苏北四市一模)已知数列,a,n,满足,a,n,=3,n,-2,f,(,n,)=,+,+,+,g,(,n,)=,f,(,n,2,)-,f,(,n,-1),n,N,*,.,(1)求证:,g,(2),;,(2)求证:当,n,3时,g,(,n,),.,证明,(1)由题意知,g,(,n,)=,+,+,+,+,a,n,=3,n,-2,g,(2)=,+,+,=,+,+,=,.,(2)用数学归纳法加以证明:,当,n,=3时,g,(3)=,+,+,+,+,=,+,+,+,+,+,+,=,+,+,+,+,=,+,+,+,+,所以当,n,=3时,结论成立.,假设,n,=,k,(,k,3)时,结论成立,即,g,(,k,),则,n,=,k,+1时,g,(,k,+1)=,g,(,k,)+,+,+,-,=,+,=,+,由,k,3可知,3,k,2,-7,k,-30,g,(,k,+1),.,所以当,n,=,k,+1时,结论也成立.,综合可得,当,n,3时,g,(,n,),.,4.,(2015江苏常州一模)设,n,个正数,a,1,a,2,a,n,满足,a,1,a,2,a,n,(,n,N,*,且,n,3).,(1)当,n,=3时,证明:,+,+,a,1,+,a,2,+,a,3,;,(2)当,n,=4时,不等式,+,+,+,a,1,+,a,2,+,a,3,+,a,4,也成立,请你将其推广到,n,(,n,N,*,且,n,3)个,正数,a,1,a,2,a,n,的情形,归纳出一般性的结论并用数学归纳法证明.,解析,(1)证明:因为,a,1,a,2,a,3,均为正实数,所以,+,+,-(,a,1,+,a,2,+,a,3,)=,+,+,+,+,=0(当且仅当,a,1,=,a,2,=,a,3,时取,等号),所以,+,+,a,1,+,a,2,+,a,3,.,(2)归纳的结论为,+,+,+,+,+,a,1,+,a,2,+,+,a,n,(,n,N,*,且,n,3).,证明如下:,记,F,n,=,+,+,+,+,+,-(,a,1,+,a,2,+,+,a,n,),当,n,=3(,n,N,*,)时,由(1)知,不等式成立;,假设当,n,=,k,(,k,N,*,且,k,3)时,不等式成立,即,F,k,=,+,+,+,+,+,-(,a,1,+,a,2,+,+,a,k,),0,则当,n,=,k,+1时,F,k,+1,=,+,+,+,+,+,+,-(,a,1,+,a,2,+,+,a,k,+,a,k,+1,),=,F,k,+,+,+,-,-,-,a,k,+1,=,F,k,+,a,k,-1,a,k,+,a,k,+1,+,(,a,k,+1,-,a,k,),0+,+,a,k,+1,+,(,a,k,+1,-,a,k,),=(,a,k,+1,-,a,k,),因为,a,k,+1,a,k,+,2,=2,所以,F,k,+1,0,所以当,n,=,k,+1时,不等式成立.,所以不等式,+,+,+,+,+,a,1,+,a,2,+,+,a,n,(,n,N,*,且,n,3)成立.,1.,(2017苏锡常镇四市高三教学情况调研(二),23)已知,f,n,(,x,)=,x,n,-,(,x,-1),n,+,+(-1),k,(,x,-,k,),n,+,+(-,1),n,(,x,-,n,),n,其中,x,R,n,N,*,k,N,k,n,.,(1)试求,f,1,(,x,),f,2,(,x,),f,3,(,x,)的值;,(2)试猜测,f,n,(,x,)关于,n,的表达式,并证明你的结论.,B,组 2015,2017年高考模拟,综合题组,(时间：45分钟 分值：50分),解答题(共50分),解析,(1),f,1,(,x,)=,x,-,(,x,-1)=,x,-,x,+1=1;,f,2,(,x,)=,x,2,-,(,x,-1),2,+,(,x,-2),2,=,x,2,-2(,x,2,-2,x,+1)+(,x,2,-4,x,+4)=2;,f,3,(,x,)=,x,3,-,(,x,-1),3,+,(,x,-2),3,-,(,x,-3),3,=,x,3,-3(,x,-1),3,+3(,x,-2),3,-(,x,-3),3,=6.,(2)猜测:,f,n,(,x,)=,n,!.,而,k,=,k,=,n,=,n,=,所以,k,=,n,.,用数学归纳法证明结论成立.,当,n,=1时,f,1,(,x,)=1,结论成立.,假设当,n,=,k,时,结论成立,即,f,k,(,x,)=,x,k,-,(,x,-1),k,+,+(-1),k,(,x,-,k,),k,=,k,!.,当,n,=,k,+1时,f,k,+1,(,x,)=,x,k,+1,-,(,x,-1),k,+1,+,+(-1),k,+1,(,x,-,k,-1),k,+1,=,x,k,+1,-,(,x,-1),k,(,x,-1)+,+(-1),k,(,x,-,k,),k,(,x,-,k,)+(-1),k,+1,(,x,-,k,-1),k,+1,=,x,x,k,-,(,x,-1),k,+,+(-1),k,(,x,-,k,),k,+,(,x,-1),k,-2,(,x,-2),k,+,+(-1),k,+1,k,(,x,-,k,),k,+(-1),k,+1,(,x,-,k,-1),k,+1,=,x,x,k,-(,+,)(,x,-1),k,+,+(-1),k,(,+,)(,x,-,k,),k,+(,k,+1)(,x,-1),k,-,(,x,-2),k,+,+(-1),k,+1,(,x,-,k,),k,+(-1),k,+1,(,x,-,k,-1),k,(,x,-,k,-1),=,x,x,k,-,(,x,-1),k,+,+(-1),k,(,x,-,k,),k,-,x,(,x,-1),k,+,+(-1),k,-1,(,x,-,k,),k,+(,k,+1)(,x,-1),k,-,(,x,-2),k,+,+(-1),k,+1,(,x,-,k,),k,+,x,(-1),k,+1,(,x,-,k,-1),k,-(,k,+1)(-1),k,+1,(,x,-,k,-1),k,=,x,x,k,-,(,x,-1),k,+,+(-1),k,(,x,-,k,),k,-,x,(,x,-1),k,+,+(-1),k,-1,(,x,-,k,),k,+(-1),k,(,x,-,k,-1),k,+(,k,+1)(,x,-1),k,-,(,x,-2),k,+,+(-1),k,-1,(,x,-,k,),k,+(-1),k,(,x,-,k,-1),k,(*).,由归纳假设知(*)式等于,x,k,!-,x,k,!+(,k,+1),k,!=(,k,+1)!.,所以当,n,=,k,+1时,结论也成立.,综合知,f,n,(,x,)=,n,!成立.,2.,(2016江苏扬州模拟,24)已知函数,f,(,x,)=2,x,-3,x,2,设数列,a,n,满足:,a,1,=,a,n,+1,=,f,(,a,n,).,(1)求证:,n,N,*,都有0,a,n,;,(2)求证:,+,+,+,4,n,+1,-4.,证明,(1)当,n,=1时,a,1,=,0,a,1,n,=1时,不等式成立.,假设当,n,=,k,(,k,N,*,)时,不等式成立,即0,a,k,则当,n,=,k,+1时,a,k,+1,=,f,(,a,k,)=2,a,k,-3,=-3,=-3,+,于是,-,a,k,+1,=3,0,a,k,03,即0,-,a,k,+1,可得0,a,k,+1,所以当,n,=,k,+1时,不等式也成立.,由可知,对任意的正整数,n,都有0,a,n,.,(2)由(1)可得,-,a,n,+1,=3,.,两边同时取以3为底的对数,可得log,3,=1+2log,3,化简为1+log,3,=2,因为1+log,3,=log,3,所以数列,是以log,3,为首项,2为公比的等比数列,1+log,3,=2,n,-1,log,3,化简求得:,-,a,n,=,=3,.,n,2时,2,n,-1,=,+,+,+,+,1+,n,-1=,n,n,=1时,2,n,-1,=1,n,N,*,时,2,n,-1,n,=3,34,n,.,+,+,+,=3(,+,+,+,),3(4,1,+4,2,+,+4,n,)=4,n,+1,-4,+,+,+,4,n,+1,-4.,3.,(2016江苏扬州中学期中,24)已知,F,n,(,x,)=,(-1),k,f,k,(,x,)(,n,N,*,).,(1)若,f,k,(,x,)=,x,k,求,F,2 015,(2)的值;,(2)若,f,k,(,x,)=,(,x,0,-1,-,n,),求证:,F,n,(,x,)=,.,解析,(1),F,n,(,x,)=,(-1),k,f,k,(,x,)=,(-,x,),k,=,(-,x,),k,1,n,-,k,=(1-,x,),n,F,2 015,(2)=-1.,(2)证明:,n,=1时,左边=1-,=,=右边.,设,n,=,m,时,对一切实数,x,(,x,0,-1,-,m,),有,=,那么,当,n,=,m,+1时,对一切实数,x,(,x,0,-1,-(,m,+1),有,=1+,+(-1),m,+1,=,+,=,-,=,-,=,=,即,n,=,m,+1时,等式成立.,故对一切正整数,n,及一切实数,x,(,x,0,-1,-,n,),有,(-1),k,=,.,4.,(2015江苏泰州二模,23)已知,f,(,x,)=(,x,2,+,x,+1),n,(,n,N,*,),g,(,x,)是关于,x,的2,n,次多项式.,(1)若,f,(,x,2,),g,(,x,)=,g,(,x,3,)恒成立,求,g,(1)和,g,(-1)的值,并写出一个满足条件的,g,(,x,)的表达式,无需证明.,(2)求证:对于任意给定的正整数,n,都存在与,x,无关的常数,a,0,a,1,a,2,a,n,使得,f,(,x,)=,a,0,(1+,x,2,n,)+,a,1,(,x,+,x,2,n,-1,)+,a,2,(,x,2,+,x,2,n,-2,)+,+,a,n,-1,(,x,n,-1,+,x,n,+1,)+,a,n,x,n,.,解析,(1)令,x,=1,则,f,(1),g,(1)=,g,(1),即,g,(1),f,(1)-1=0,因为,f,(1)-1=3,n,-1,0,所以,g,(1)=0.,令,x,=-1,则,f,(-1),2,g,(-1)=,g,(-1),3,即,f,(1),g,(-1)=,g,(-1),即,g,(-1),f,(1)-1=0,因为,f,(1)-1=3,n,-1,0,所以,g,(-1)=0.,满足条件的,g,(,x,)的表达式:,g,(,x,)=(,x,2,-1),n,(,n,N,*,).,(2)当,n,=1时,f,(,x,)=,x,2,+,x,+1=(,x,2,+1)+,x,故存在常数,a,0,=1,a,1,=1,使得,f,(,x,)=,a,0,(1+,x,2,)+,a,1,x,.,假设当,n,=,k,(,k,N,*,)时,存在与,x,无关的常数,a,0,a,1,a,2,a,k,使得,f,(,x,)=,a,0,(1+,x,2,k,)+,a,1,(,x,+,x,2,k,-1,)+,a,2,(,x,2,+,x,2,k,-2,)+,+,a,k,-1,(,x,k,-1,+,x,k,+1,)+,a,k,x,k,即(,x,2,+,x,+1),k,=,a,0,(1+,x,2,k,)+,a,1,(,x,+,x,2,k,-1,)+,a,2,(,x,2,+,x,2,k,-2,)+,+,a,k,-1,(,x,k,-1,+,x,k,+1,)+,a,k,x,k,则当,n,=,k,+1时,f,(,x,)=(,x,2,+,x,+1),k,+1,=(,x,2,+,x,+1)(,x,2,+,x,+1),k,=(,x,2,+,x,+1),a,0,(1+,x,2,k,)+,a,1,(,x,+,x,2,k,-1,)+,+,a,k,-1,(,x,k,-1,+,x,k,+1,)+,a,k,x,k,=(,a,0,+,a,1,x,+,+,a,k,-1,x,k,-1,+,a,k,x,k,+,a,k,-1,x,k,+1,+,+,a,1,x,2,k,-1,+,a,0,x,2,k,),+(,a,0,x,+,a,1,x,2,+,+,a,k,-1,x,k,+,a,k,x,k,+1,+,a,k,-1,x,k,+2,+,+,a,1,x,2,k,+,a,0,x,2,k,+1,),+(,a,0,x,2,+,a,1,x,3,+,+,a,k,-1,x,k,+1,+,a,k,x,k,+2,+,a,k,-1,x,k,+3,+,+,a,1,x,2,k,+1,+,a,0,x,2,k,+2,),=,a,0,+(,a,1,+,a,0,),x,+(,a,2,+,a,1,+,a,0,),x,2,+(,a,3,+,a,2,+,a,1,),x,3,+,+(,a,k,-1,+,a,k,-2,+,a,k,-3,),x,k,-1,+(,a,k,+,a,k,-1,+,a,k,-2,),x,k,+(2,a,k,-1,+,a,k,),x,k,+1,+(,a,k,+,a,k,-1,+,a,k,-,2,),x,k,+2,+,+(,a,3,+,a,2,+,a,1,),x,2,k,-1,+(,a,2,+,a,1,+,a,0,),x,2,k,+(,a,1,+,a,0,),x,2,k,+1,+,a,0,x,2,k,+2,=,a,0,(1+,x,2,k,+2,)+(,a,1,+,a,0,)(,x,+,x,2,k,+1,)+(,a,2,+,a,1,+,a,0,)(,x,2,+,x,2,k,)+,+(,a,k,+,a,k,-1,+,a,k,-2,)(,x,k,+,x,k,+2,)+(2,a,k,-1,+,a,k,),x,k,+1,令,a,0,=,a,0,a,1,=,a,0,+,a,1,a,m,=,a,m,-2,+,a,m,-1,+,a,m,(2,m,k,),a,k,+1,=2,a,k,-1,+,a,k,故存在与,x,无关的常数,a,0,a,1,a,2,a,k,a,k,+1,使得,f,(,x,)=,a,0,(1+,x,2,k,+2,)+,a,1,(,x,+,x,2,k,+1,)+,a,2,(,x,2,+,x,2,k,)+,+,a,k,(,x,k,+,x,k,+2,),+,a,k,+1,x,k,+1,.,综上所述,对于任意给定的正整数,n,都存在与,x,无关的常数,a,0,a,1,a,2,a,n,使得,f,(,x,)=,a,0,(1+,x,2,n,)+,a,1,(,x,+,x,2,n,-,1,)+,a,2,(,x,2,+,x,2,n,-2,)+,+,a,n,-1,(,x,n,-1,+,x,n,+1,)+,a,n,x,n,.,