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第三章 数学归纳法与贝努利不等式3 3.1 1数学归纳法原理数学归纳法原理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航1.了解数学归纳法的原理.2.了解数学归纳法的应用范围.3.会用数学归纳法证明一些简单问题.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航1.归纳法由有限多个个别的特殊事例得出一般结论的推理方法,通常称为归纳法.名师点拨根据推理过程中考察的对象是涉及事物的一部分还是全部,归纳法分为不完全归纳法和完全归纳法.(1)不完全归纳法是根据事物的部分(而不是全部)特例得到一般结论的推理方法.不完全归纳法所得到的命题不一定是成立的,但它是一种重要的思考问题的方法,是研究数学问题的一把钥匙,是发现数学规律的一种重要手段.用不完全归纳法发现规律,用数学归纳法证明是解决问题的一种重要途径.(2)完全归纳法是一种在研究了事物的所有(有限种)特殊情况后得出一般结论的推理方法,又叫枚举法.与不完全归纳法不同,用完全归纳法得出的结论是可靠的.通常在事物包括的特殊情况不多时,采用完全归纳法.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航【做一做1-2】 从1=1,1-4=-(1+2),1-4+9=1+2+3,猜想第n个式子为.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航2.数学归纳法一般地,当要证明一个命题对于不小于某正数n0的所有正整数n都成立时,可以用以下两个步骤:(1)证明当n=n0时命题成立;(2)假设当n=k(kN,且kn0)时命题成立,证明当n=k+1时命题也成立.完成两个步骤后,就可以断定命题对于不小于n0的所有正整数都成立,这种证明方法称为数学归纳法.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航名师点拨1.这两个步骤缺一不可,只完成步骤(1)而缺少步骤(2),就作出判断可能得出不正确的结论.因为单靠步骤(1),无法递推下去,即n取n0以后的数时命题是否正确,我们无法判定.同样,只有步骤(2)而缺少步骤(1),也可能得出不正确的结论.缺少步骤(1)这个基础,假设就失去了成立的前提,步骤(2)也就没有意义了.2.用数学归纳法证明有关问题的关键在第二步,即n=k+1时为什么成立?n=k+1时成立是利用假设n=k时成立,根据有关的定理、定义、公式、性质等数学结论推证出n=k+1时命题成立,而不是直接代入,否则n=k+1时也成假设了,命题并没有得到证明.3.用数学归纳法可证明有关的正整数问题,但并不是所有的正整数问题都用数学归纳法证明,学习时要具体问题具体分析.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航【做一做2-1】 下列说法中不正确的是()A.数学归纳法中的两个步骤相互依存,缺一不可B.数学归纳法证明的是与正整数有关的命题C.数学归纳法证明的第一步是递推的基础,第二步是递推的依据D.数学归纳法中第一步必须从n=1开始答案:D目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航故当n=k+1时,不等式成立.上述的证明过程中,不正确的一步的序号为.解析:在(2)中,由n=k到n=k+1的证明,没有用上归纳假设,故(2)错误.答案:(2)目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航1.为什么数学归纳法能够证明无限多正整数都成立的问题呢?剖析:这是因为第一步首先验证了n取第一个值n0时命题成立,这样假设就有了存在的基础.假设当n=k时命题成立,根据假设和合理推证,证明出当n=k+1时命题也成立.这实质上是证明了一种循环.如验证了当n0=1时命题成立,又证明了当n=k+1时命题也成立,这就一定有当n=2时命题成立,当n=2时命题成立,则当n=3时命题也成立;当n=3时命题成立,则当n=4时命题也成立.如此反复,以至无穷.对所有nn0的正整数命题就都成立了.数学归纳法非常巧妙地解决了一种无限多的正整数问题,这就是数学方法的神奇.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航2.什么时候可以运用数学归纳法证明,证明时n0是否一定要为1?剖析:数学归纳法一般被用于证明某些涉及正整数n的命题,n可取无限多值,但不能简单地说所有涉及正整数n的命题都可以用数学归纳法证明,例如用数学归纳法证明 (nN*)的单调性就难以实现,一般说来,从n=k到n=k+1时,若问题中存在可利用的递推关系,则使用数学归纳法就较简单,否则使用数学归纳法就有困难.在运用数学归纳法时,要注意起点n并非一定取1,也可能取2等值,要看清题目,比如证明凸n边形的内角和f(n)=(n-2)180,这里面的n应不小于3,即n3,第一个值n0=3.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四用数学归纳法证明恒等式【例1】 用数学归纳法证明:分析:用数学归纳法证明一个与正整数有关的命题的关键是第二步,要注意当n=k+1时等式两边的式子与n=k时等式两边的式子的联系,增加了哪些项,减少了哪些项,问题就会顺利解决.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型四题型三用数学归纳法证明整除性问题【例2】 求证:an+1+(a+1)2n-1能被a2+a+1整除,nN*.分析:对于多项式A,B,如果A=BC,C也是多项式,那么A能被B整除.若A,B都能被C整除,则A+B,A-B也能被C整除.证明:(1)当n=1时,a1+1+(a+1)21-1=a2+a+1,命题显然成立.(2)假设当n=k(kN*,且k1)时,ak+1+(a+1)2k-1能被a2+a+1整除,则当n=k+1时,ak+2+(a+1)2k+1=aak+1+(a+1)2(a+1)2k-1=aak+1+(a+1)2k-1+(a+1)2(a+1)2k-1-a(a+1)2k-1=aak+1+(a+1)2k-1+(a2+a+1)(a+1)2k-1.由归纳假设,得上式中的两项均能被a2+a+1整除,故当n=k+1时命题成立.根据(1)(2)可知,对一切nN*,命题成立.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型四题型三反思证明整除性问题的关键是“凑项”,采用增项、减项、拆项、因式分解等手段,凑出当n=k时的情形,从而利用归纳假设使问题得证.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四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个部分(nN*).分析:因为f(n)为n个圆把平面分割成的区域数,那么再有一个圆和这n个圆相交,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,所以增加一个圆后,平面分成的区域数增加2n,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SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四证明:(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n=1时命题成立.(2)假设n=k(kN*,且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,在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.故当n=k+1时,命题成立.根据(1)(2)可知,对一切nN*,命题成立.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四反思对于用数学归纳法证明几何问题,可以先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,再去证明.也可以用“递推”的办法,比如本题,当n=k+1时的结果已知道:f(k+1)=(k+1)2-(k+1)+2,用f(k+1)-f(k)就可得到增加的部分,然后从有限的情况来理解如何增加的,也就好理解了.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四易错辨析易错点:在应用数学归纳法证明有关问题时,两步缺一不可,且最易出错的地方是在第二步证明中未用归纳假设.【例4】 已知在数列an中,a1=3,其前n项和Sn满足Sn=6-2an+1,计算a2,a3,a4,然后猜想出an的表达式,并用数学归纳法证明你的结论.错解:当n2时,an=Sn-Sn-1=6-2an+1-(6-2an)目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四错因分析:本题在证明时出现的主要错误是未用归纳假设.目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航题型一题型二题型三题型四目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航1 2 3 4 51下列代数式中,nN*,则可能被13整除的是()A.n3+5nB.34n+1+52n+1C.62n-1+1D.42n+1+3n+2解析:当n=1时,只有D项能被13整除.答案:D目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航1 2 3 4 52若凸n边形有f(n)条对角线,则凸(n+1)边形的对角线的条数f(n+1)为()A.f(n)+n+1B.f(n)+nC.f(n)+n-1D.f(n)+n-2解析:从凸n边形到凸(n+1)边形,对角线增加了(n-1)条.答案:C目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航1 2 3 4 53下列四个判断中,正确的是()A.式子1+k+k2+kn(nN*),当n=1时为1B.式子1+k+k2+kn-1(nN*),当n=1时为1+k解析:对于选项A,n=1时,式子应为1+k;选项B中,n=1时,式子应为1;答案:C 目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航1 2 3 4 54已知在数列an中,a1=1,a2=2,an+1=2an+an-1(nN*),用数学归纳法证明a4n能被4整除,假设a4k能被4整除,则下一步证明.答案:a4k+4能被4整除目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理DIANLITOUXI典例透析SUITANGLIANXI随堂演练ZHONGNANJUJIAO重难聚焦ZHISHISHULI知识梳理目标导航1 2 3 4 55某同学用数学归纳法证明等式1+2+22+2n-1=2n-1的过程如下:(1)当n=1时,左边=1,右边=1,等式成立;(2)假设当n=k(kN*,且k1)时,等式成立,即1+2+22+2k-1=2k-1;即当n=k+1时等式成立.根据(1)(2)可知,对任意正整数n等式成立.以上证明过程的错误是.答案:第(2)步未用归纳假设
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