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3数学归纳法与贝努利不等式3 3.1 1数学归纳法目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理1.理解数学归纳法的原理和实质.2.掌握用数学归纳法证明与正整数有关的命题的两个步骤,并能灵活运用.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理对数学归纳法的理解(1)数学归纳法原理:数学归纳法原理是设有一个关于正整数n的命题,若当n取第1个值n0时该命题成立,又在假设当n取第k个值时该命题成立后可以推出n取第k+1个值时该命题成立,则该命题对一切自然数nn0都成立.(2)数学归纳法:数学归纳法可以用于证明与正整数有关的命题.证明需要经过两个步骤:验证当n取第一个值n0(如n0=1或2等)时命题正确.假设当n=k时(kN+,kn0)命题正确,证明当n=k+1时命题也正确.在完成了上述两个步骤之后,就可以断定命题对于从n0开始的所有正整数都正确.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理【做一做1】 在用数学归纳法证明多边形内角和定理时,第一步应检验()A.当n=1时成立B.当n=2时成立C.当n=3时成立D.当n=4时成立解析:多边形中至少有三条边,故应先验证当n=3时成立.答案:C目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理A.当n=k+1时等式成立B.当n=k+2时等式成立C.当n=2k+2时等式成立D.当n=2(k+2)时等式成立解析:因为已假设当n=k(k2,且k为偶数)时命题为真,即当n=k+2时命题为真.而选项中n=k+1为奇数,n=2k+2和n=2(k+2)均不满足递推关系,所以只有n=k+2满足条件.答案:B目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理A.2kB.2k-1C.2k-1D.2k+1答案:A目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三题型一 用数学归纳法证明恒等问题 分析:在证明时,要严格按数学归纳法的步骤进行,并要特别注意当n=k+1时等式两边的式子,与当n=k时等式两边的式子之间的联系,明确增加了哪些项,减少了哪些项.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三反思在解本题时,当由n=k到n=k+1时,等式的左边增加了一项,这里容易因忽略而出错.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三【变式训练1】 用数学归纳法证明: 目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三题型二 用数学归纳法证明整除问题【例2】用数学归纳法证明:n3+5n(nN+)能被6整除.分析:这是一个与整除有关的命题,它涉及全体正整数,第一步应证明当n=1时成立,第二步应明确目标,在假设k3+5k能被6整除的前提下,证明(k+1)3+5(k+1)也能被6整除.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三证明:(1)当n=1时,n3+5n=6显然能被6整除,命题成立.(2)假设当n=k(kN+,且k1)时,命题成立,即k3+5k能被6整除.则当n=k+1时,(k+1)3+5(k+1)=k3+3k2+3k+1+5k+5=(k3+5k)+3k2+3k+6=(k3+5k)+3k(k+1)+6.由假设知k3+5k能够被6整除,而k(k+1)是偶数,故3k(k+1)能够被6整除,从而(k3+5k)+3k(k+1)+6,即(k+1)3+5(k+1)能够被6整除.因此,当n=k+1时,命题也成立.由(1)(2)知,命题对一切正整数成立,即n3+5n(nN+)能被6整除.反思用数学归纳法证明有关整除性问题的关键是寻找f(k+1)与f(k)之间的递推关系,基本策略就是“往后退”,从f(k+1)中将f(k)分离出来.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三【变式训练2】 用数学归纳法证明:1-(3+x)n(nN+)能被x+2整除.证明:(1)当n=1时,1-(3+x)=-(x+2),能被x+2整除,命题成立.(2)假设当n=k(kN+,且k1)时,1-(3+x)n能被x+2整除,则可设1-(3+x)k=(x+2)f(x) (f(x)为k-1次多项式).则当n=k+1时,1-(3+x)k+1=1-(3+x)(3+x)k=1-(3+x)1-(x+2)f(x)=1-(3+x)+(x+2)(3+x)f(x)=-(x+2)+(x+2)(3+x)f(x)=(x+2)-1+(3+x)f(x),能被x+2整除,即当n=k+1时命题也成立.由(1)(2)可知,对nN+,1-(3+x)n能被x+2整除.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三题型三 利用数学归纳法证明几何问题【例3】 平面内有n个圆,任意两个圆都相交于两点,任意三个圆不相交于同一点,求证:这n个圆将平面分成f(n)=n2-n+2(nN+)个部分.分析:因为f(n)为n个圆把平面分割成的区域数,如果再有一个圆和这n个圆相交,那么就有2n个交点,这些交点将增加的这个圆分成2n段弧,且每一段弧又将原来的平面区域一分为二,因此,增加一个圆后,平面分成的区域数增加2n个,即f(n+1)=f(n)+2n.有了上述关系,数学归纳法的第二步证明可迎刃而解.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三证明:(1)当n=1时,一个圆将平面分成两个部分,且f(1)=1-1+2=2,所以当n=1时命题成立.(2)假设当n=k(kN+,k1)时命题成立,即k个圆把平面分成f(k)=k2-k+2个部分,则当n=k+1时,从(k+1)个圆中任取一个圆O,剩下的k个圆将平面分成f(k)个部分,而圆O与k个圆有2k个交点,这2k个交点将圆O分成2k段弧,每段弧将原平面一分为二,故得f(k+1)=f(k)+2k=k2-k+2+2k=(k+1)2-(k+1)+2.所以当n=k+1时,命题也成立.综合(1)(2)可知,对一切nN+命题成立.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三反思对于几何问题的证明,可以从有限情形中归纳出一个变化的过程,或者说体会出是怎样变化的,然后再去证明,也可以用“递推”的办法.目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理题型一题型二题型三目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理12341用数学归纳法证明1+2+(2n+1)=(n+1)(2n+1)时,在验证n=1成立时,左边所得的代数式是()A.1B.1+3C.1+2+3D.1+2+3+4答案:C目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理1234答案:B 目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理12343用数学归纳法证明关于n的恒等式,当n=k时,表达式为14+27+k(3k+1)=k(k+1)2,则当n=k+1时,表达式为.答案:14+27+k(3k+1)+(k+1)(3k+4)=(k+1)(k+2)2目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGYANLIAN随堂演练ZHISHISHULI知识梳理1234
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