高中數(shù)學(xué) 第三章 數(shù)學(xué)歸納法與貝努利不等式 3.1 數(shù)學(xué)歸納法原理課件 新人教B版選修45

《高中數(shù)學(xué) 第三章 數(shù)學(xué)歸納法與貝努利不等式 3.1 數(shù)學(xué)歸納法原理課件 新人教B版選修45》由會員分享，可在線閱讀，更多相關(guān)《高中數(shù)學(xué) 第三章 數(shù)學(xué)歸納法與貝努利不等式 3.1 數(shù)學(xué)歸納法原理課件 新人教B版選修45（27頁珍藏版）》請?jiān)谘b配圖網(wǎng)上搜索。

1、第三章 數(shù)學(xué)歸納法與貝努利不等式3 3.1 1數(shù)學(xué)歸納法原理數(shù)學(xué)歸納法原理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZH

2、ISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1.了解數(shù)學(xué)歸納法的原理.2.了解數(shù)學(xué)歸納法的應(yīng)用范圍.3.會用數(shù)學(xué)歸納法證明一些簡單問題.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂

3、演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1.歸納法由有限多個(gè)個(gè)別的特殊事例得出一般結(jié)論的推理方法,通常稱為歸納法.名師點(diǎn)撥根據(jù)推理過程中考察的對象是涉及事物的一部分還是全部,歸納法分為不完全歸納法和完全歸納法.(1)不完全歸納法是根據(jù)事物的部分(而不是全部)特例得到一般結(jié)論的推理方法.不完全歸

4、納法所得到的命題不一定是成立的,但它是一種重要的思考問題的方法,是研究數(shù)學(xué)問題的一把鑰匙,是發(fā)現(xiàn)數(shù)學(xué)規(guī)律的一種重要手段.用不完全歸納法發(fā)現(xiàn)規(guī)律,用數(shù)學(xué)歸納法證明是解決問題的一種重要途徑.(2)完全歸納法是一種在研究了事物的所有(有限種)特殊情況后得出一般結(jié)論的推理方法,又叫枚舉法.與不完全歸納法不同,用完全歸納法得出的結(jié)論是可靠的.通常在事物包括的特殊情況不多時(shí),采用完全歸納法.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHON

5、GNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航【做一做1-2】 從1=1,1-4=-(1+2),1-4+9=1+2+3,猜想第n個(gè)式子為.目標(biāo)導(dǎo)航DIANLITOUXI典例透

6、析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJU

7、JIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航2.數(shù)學(xué)歸納法一般地,當(dāng)要證明一個(gè)命題對于不小于某正數(shù)n0的所有正整數(shù)n都成立時(shí),可以用以下兩個(gè)步驟:(1)證明當(dāng)n=n0時(shí)命題成立;(2)假設(shè)當(dāng)n=k(kN,且kn0)時(shí)命題成立,證明當(dāng)n=k+1時(shí)命題也成立.完成兩個(gè)步驟后,就可以斷定命題對于不小于n0的所有正整數(shù)都成立,這種證明方法稱為數(shù)學(xué)歸納法.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難

8、聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航名師點(diǎn)撥1.這兩個(gè)步驟缺一不可,只完成步驟(1)而缺少步驟(2),就作出判斷可能得出不正確的結(jié)論.因?yàn)閱慰坎襟E(1),無法遞推下去,即n取n0以后的數(shù)時(shí)

9、命題是否正確,我們無法判定.同樣,只有步驟(2)而缺少步驟(1),也可能得出不正確的結(jié)論.缺少步驟(1)這個(gè)基礎(chǔ),假設(shè)就失去了成立的前提,步驟(2)也就沒有意義了.2.用數(shù)學(xué)歸納法證明有關(guān)問題的關(guān)鍵在第二步,即n=k+1時(shí)為什么成立?n=k+1時(shí)成立是利用假設(shè)n=k時(shí)成立,根據(jù)有關(guān)的定理、定義、公式、性質(zhì)等數(shù)學(xué)結(jié)論推證出n=k+1時(shí)命題成立,而不是直接代入,否則n=k+1時(shí)也成假設(shè)了,命題并沒有得到證明.3.用數(shù)學(xué)歸納法可證明有關(guān)的正整數(shù)問題,但并不是所有的正整數(shù)問題都用數(shù)學(xué)歸納法證明,學(xué)習(xí)時(shí)要具體問題具體分析.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONG

10、NANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理

11、目標(biāo)導(dǎo)航【做一做2-1】 下列說法中不正確的是()A.數(shù)學(xué)歸納法中的兩個(gè)步驟相互依存,缺一不可B.數(shù)學(xué)歸納法證明的是與正整數(shù)有關(guān)的命題C.數(shù)學(xué)歸納法證明的第一步是遞推的基礎(chǔ),第二步是遞推的依據(jù)D.數(shù)學(xué)歸納法中第一步必須從n=1開始答案:D目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNA

12、NJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航故當(dāng)n=k+1時(shí),不等式成立.上述的證明過程中,不正確的一步的序號為.解析:在(2)中,由n=k到n=k+1的證明,沒有用上歸納假設(shè),故(2)錯(cuò)誤.答案:(2)目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZH

13、ISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1.為什么數(shù)學(xué)歸納法能

14、夠證明無限多正整數(shù)都成立的問題呢?剖析:這是因?yàn)榈谝徊绞紫闰?yàn)證了n取第一個(gè)值n0時(shí)命題成立,這樣假設(shè)就有了存在的基礎(chǔ).假設(shè)當(dāng)n=k時(shí)命題成立,根據(jù)假設(shè)和合理推證,證明出當(dāng)n=k+1時(shí)命題也成立.這實(shí)質(zhì)上是證明了一種循環(huán).如驗(yàn)證了當(dāng)n0=1時(shí)命題成立,又證明了當(dāng)n=k+1時(shí)命題也成立,這就一定有當(dāng)n=2時(shí)命題成立,當(dāng)n=2時(shí)命題成立,則當(dāng)n=3時(shí)命題也成立;當(dāng)n=3時(shí)命題成立,則當(dāng)n=4時(shí)命題也成立.如此反復(fù),以至無窮.對所有nn0的正整數(shù)命題就都成立了.數(shù)學(xué)歸納法非常巧妙地解決了一種無限多的正整數(shù)問題,這就是數(shù)學(xué)方法的神奇.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演

15、練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHU

16、LI知識梳理目標(biāo)導(dǎo)航2.什么時(shí)候可以運(yùn)用數(shù)學(xué)歸納法證明,證明時(shí)n0是否一定要為1?剖析:數(shù)學(xué)歸納法一般被用于證明某些涉及正整數(shù)n的命題,n可取無限多值,但不能簡單地說所有涉及正整數(shù)n的命題都可以用數(shù)學(xué)歸納法證明,例如用數(shù)學(xué)歸納法證明 (nN*)的單調(diào)性就難以實(shí)現(xiàn),一般說來,從n=k到n=k+1時(shí),若問題中存在可利用的遞推關(guān)系,則使用數(shù)學(xué)歸納法就較簡單,否則使用數(shù)學(xué)歸納法就有困難.在運(yùn)用數(shù)學(xué)歸納法時(shí),要注意起點(diǎn)n并非一定取1,也可能取2等值,要看清題目,比如證明凸n邊形的內(nèi)角和f(n)=(n-2)180,這里面的n應(yīng)不小于3,即n3,第一個(gè)值n0=3.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUI

17、TANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO

18、重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四用數(shù)學(xué)歸納法證明恒等式【例1】 用數(shù)學(xué)歸納法證明:分析:用數(shù)學(xué)歸納法證明一個(gè)與正整數(shù)有關(guān)的命題的關(guān)鍵是第二步,要注意當(dāng)n=k+1時(shí)等式兩邊的式子與n=k時(shí)等式兩邊的式子的聯(lián)系,增加了哪些項(xiàng),減少了哪些項(xiàng),問題就會順利解決.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典

19、例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLI

20、ANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHON

21、GNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳

22、理目標(biāo)導(dǎo)航題型一題型二題型四題型三用數(shù)學(xué)歸納法證明整除性問題【例2】 求證:an+1+(a+1)2n-1能被a2+a+1整除,nN*.分析:對于多項(xiàng)式A,B,如果A=BC,C也是多項(xiàng)式,那么A能被B整除.若A,B都能被C整除,則A+B,A-B也能被C整除.證明:(1)當(dāng)n=1時(shí),a1+1+(a+1)21-1=a2+a+1,命題顯然成立.(2)假設(shè)當(dāng)n=k(kN*,且k1)時(shí),ak+1+(a+1)2k-1能被a2+a+1整除,則當(dāng)n=k+1時(shí),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

23、-1=aak+1+(a+1)2k-1+(a2+a+1)(a+1)2k-1.由歸納假設(shè),得上式中的兩項(xiàng)均能被a2+a+1整除,故當(dāng)n=k+1時(shí)命題成立.根據(jù)(1)(2)可知,對一切nN*,命題成立.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHUL

24、I知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型四題型三反思證明整除性問題的關(guān)鍵是“湊項(xiàng)”,采用增項(xiàng)、減項(xiàng)、拆項(xiàng)、因式分解等手段,湊出當(dāng)n=k時(shí)的情形,從而利用歸納假設(shè)使問題得證.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUX

25、I典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四用數(shù)學(xué)歸納法證明幾何問題【例3】 平面內(nèi)有n個(gè)圓,任

26、意兩個(gè)圓都相交于兩點(diǎn),任意三個(gè)圓不相交于同一點(diǎn),求證:這n個(gè)圓將平面分成f(n)=n2-n+2個(gè)部分(nN*).分析:因?yàn)閒(n)為n個(gè)圓把平面分割成的區(qū)域數(shù),那么再有一個(gè)圓和這n個(gè)圓相交,就有2n個(gè)交點(diǎn),這些交點(diǎn)將增加的這個(gè)圓分成2n段弧,且每一段弧又將原來的平面區(qū)域一分為二,所以增加一個(gè)圓后,平面分成的區(qū)域數(shù)增加2n,即f(n+1)=f(n)+2n.有了上述關(guān)系,數(shù)學(xué)歸納法的第二步證明可迎刃而解.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIA

27、NXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四證明:(1)當(dāng)n=1時(shí),一個(gè)圓將平面分成兩個(gè)部分,且f(1)=1-1+2=2,所以

28、n=1時(shí)命題成立.(2)假設(shè)n=k(kN*,且k1)時(shí)命題成立,即k個(gè)圓把平面分成f(k)=k2-k+2個(gè)部分,則當(dāng)n=k+1時(shí),在k+1個(gè)圓中任取一個(gè)圓O,剩下的k個(gè)圓將平面分成f(k)個(gè)部分,而圓O與k個(gè)圓有2k個(gè)交點(diǎn),這2k個(gè)點(diǎn)將圓O分成2k段弧,每段弧將原平面一分為二,故得f(k+1)=f(k)+2k=k2-k+2+2k=(k+1)2-(k+1)+2.故當(dāng)n=k+1時(shí),命題成立.根據(jù)(1)(2)可知,對一切nN*,命題成立.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOU

29、XI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四反思對于用數(shù)學(xué)歸納法證明幾何問題,可以先從有限情形

30、中歸納出一個(gè)變化的過程,或者說體會出是怎樣變化的,再去證明.也可以用“遞推”的辦法,比如本題,當(dāng)n=k+1時(shí)的結(jié)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,然后從有限的情況來理解如何增加的,也就好理解了.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨

31、堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四易錯(cuò)辨析易錯(cuò)點(diǎn):在應(yīng)用數(shù)學(xué)歸納法證明有關(guān)問題時(shí),兩步缺一不可,且最易出錯(cuò)的地方是在第二步證明中未用歸納假設(shè).【例4】 已知在數(shù)列an中,a1=3,其前n項(xiàng)和Sn滿足Sn=6-2an+1,計(jì)算a2,a3,a4,然后猜想出an的表達(dá)

32、式,并用數(shù)學(xué)歸納法證明你的結(jié)論.錯(cuò)解:當(dāng)n2時(shí),an=Sn-Sn-1=6-2an+1-(6-2an)目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNA

33、NJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四錯(cuò)因分析:本題在證明時(shí)出現(xiàn)的主要錯(cuò)誤是未用歸納假設(shè).目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIA

34、NXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONG

35、NANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 51下列代數(shù)式中,nN*,則可能被13整除的是()A.n3+5nB.34n+1+52n+1C.62n-1+1D.4

36、2n+1+3n+2解析:當(dāng)n=1時(shí),只有D項(xiàng)能被13整除.答案:D目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHI

37、SHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 52若凸n邊形有f(n)條對角線,則凸(n+1)邊形的對角線的條數(shù)f(n+1)為()A.f(n)+n+1B.f(n)+nC.f(n)+n-1D.f(n)+n-2解析:從凸n邊形到凸(n+1)邊形,對角線增加了(n-1)條.答案:C目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIAN

38、XI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 53下列四個(gè)判斷中,正確的是()A.式子1+k+k2+kn(nN*),當(dāng)n=1時(shí)為1B.式子

39、1+k+k2+kn-1(nN*),當(dāng)n=1時(shí)為1+k解析:對于選項(xiàng)A,n=1時(shí),式子應(yīng)為1+k;選項(xiàng)B中,n=1時(shí),式子應(yīng)為1;答案:C 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITA

40、NGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 54已知在數(shù)列an中,a1=1,a2=2,an+1=2an+an-1(nN*),用數(shù)學(xué)歸納法證明a4n能被4整除,假設(shè)a4k能被4整除,則下一步證明.答案:a4k+4能被4整除目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SU

41、ITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 55某同學(xué)用數(shù)學(xué)歸納法證明等式1+2+22+2n-1=2n-1的過程如下:(1)當(dāng)n=1時(shí),左邊=1,右邊=1,等式成立;(2)假設(shè)當(dāng)n=k(kN*,且k1)時(shí),等式成立,即1+2+22+2k-1=2k-1;即當(dāng)n=k+1時(shí)等式成立.根據(jù)(1)(2)可知,對任意正整數(shù)n等式成立.以上證明過程的錯(cuò)誤是.答案:第(2)步未用歸納假設(shè)