高中數(shù)學(xué) 第二章 幾個(gè)重要的不等式 2.3 數(shù)學(xué)歸納法與貝努利不等式 2.3.1 數(shù)學(xué)歸納法課件 北師大版選修45

《高中數(shù)學(xué) 第二章 幾個(gè)重要的不等式 2.3 數(shù)學(xué)歸納法與貝努利不等式 2.3.1 數(shù)學(xué)歸納法課件 北師大版選修45》由會(huì)員分享，可在線閱讀，更多相關(guān)《高中數(shù)學(xué) 第二章 幾個(gè)重要的不等式 2.3 數(shù)學(xué)歸納法與貝努利不等式 2.3.1 數(shù)學(xué)歸納法課件 北師大版選修45（21頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、3數(shù)學(xué)歸納法與貝努利不等式3 3.1 1數(shù)學(xué)歸納法目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1.理解數(shù)學(xué)歸納法的原理和實(shí)質(zhì).2.掌握用數(shù)學(xué)歸納法證明與正整數(shù)有關(guān)的命題的兩個(gè)步驟,并能靈活運(yùn)用.目標(biāo)導(dǎo)航DIANLITOUXI典例

2、透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理對(duì)數(shù)學(xué)歸納法的理解(1)數(shù)學(xué)歸納法原理:數(shù)學(xué)歸納法原理是設(shè)有一個(gè)關(guān)于正整數(shù)n的命題,若當(dāng)n取第1個(gè)值n0時(shí)該命題成立,又在假設(shè)當(dāng)n取第k個(gè)值時(shí)該命題成立后可以推出n取第k+1個(gè)值時(shí)該命題成立,則該命題對(duì)一切自然數(shù)nn0

3、都成立.(2)數(shù)學(xué)歸納法:數(shù)學(xué)歸納法可以用于證明與正整數(shù)有關(guān)的命題.證明需要經(jīng)過(guò)兩個(gè)步驟:驗(yàn)證當(dāng)n取第一個(gè)值n0(如n0=1或2等)時(shí)命題正確.假設(shè)當(dāng)n=k時(shí)(kN+,kn0)命題正確,證明當(dāng)n=k+1時(shí)命題也正確.在完成了上述兩個(gè)步驟之后,就可以斷定命題對(duì)于從n0開(kāi)始的所有正整數(shù)都正確.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳

4、理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理【做一做1】 在用數(shù)學(xué)歸納法證明多邊形內(nèi)角和定理時(shí),第一步應(yīng)檢驗(yàn)()A.當(dāng)n=1時(shí)成立B.當(dāng)n=2時(shí)成立C.當(dāng)n=3時(shí)成立D.當(dāng)n=4時(shí)成立解析:多邊形中至少有三條邊,故應(yīng)先驗(yàn)證當(dāng)n=3時(shí)成立.答案:C目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISH

5、ISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理A.當(dāng)n=k+1時(shí)等式成立B.當(dāng)n=k+2時(shí)等式成立C.當(dāng)n=2k+2時(shí)等式成立D.當(dāng)n=2(k+2)時(shí)等式成立解析:因?yàn)橐鸭僭O(shè)當(dāng)n=k(k2,且k為偶數(shù))時(shí)命題為真,即當(dāng)n=k+2時(shí)命題為真.而選項(xiàng)中n=k+1為奇數(shù),n=2k+2和n=2(k+2)均不滿足遞推關(guān)系,所以只有n=k+2滿足條件.答案:B目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨

6、堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理A.2kB.2k-1C.2k-1D.2k+1答案:A目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI

7、知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三題型一 用數(shù)學(xué)歸納法證明恒等問(wèn)題 分析:在證明時(shí),要嚴(yán)格按數(shù)學(xué)歸納法的步驟進(jìn)行,并要特別注意當(dāng)n=k+1時(shí)等式兩邊的式子,與當(dāng)n=k時(shí)等式兩邊的式子之間的聯(lián)系,明確增加了哪些項(xiàng),減少了哪些項(xiàng).目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂

8、演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三反思在解本題時(shí),當(dāng)由n=k到n=k+1時(shí),等式的左邊增加了一項(xiàng),這里容易因忽略而出錯(cuò).目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUIT

9、ANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三【變式訓(xùn)練1】 用數(shù)學(xué)歸納法證明: 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三題型二 用數(shù)學(xué)歸納法證明整除問(wèn)題【例2】用數(shù)學(xué)歸納法證

10、明:n3+5n(nN+)能被6整除.分析:這是一個(gè)與整除有關(guān)的命題,它涉及全體正整數(shù),第一步應(yīng)證明當(dāng)n=1時(shí)成立,第二步應(yīng)明確目標(biāo),在假設(shè)k3+5k能被6整除的前提下,證明(k+1)3+5(k+1)也能被6整除.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練Z

11、HISHISHULI知識(shí)梳理題型一題型二題型三證明:(1)當(dāng)n=1時(shí),n3+5n=6顯然能被6整除,命題成立.(2)假設(shè)當(dāng)n=k(kN+,且k1)時(shí),命題成立,即k3+5k能被6整除.則當(dāng)n=k+1時(shí),(k+1)3+5(k+1)=k3+3k2+3k+1+5k+5=(k3+5k)+3k2+3k+6=(k3+5k)+3k(k+1)+6.由假設(shè)知k3+5k能夠被6整除,而k(k+1)是偶數(shù),故3k(k+1)能夠被6整除,從而(k3+5k)+3k(k+1)+6,即(k+1)3+5(k+1)能夠被6整除.因此,當(dāng)n=k+1時(shí),命題也成立.由(1)(2)知,命題對(duì)一切正整數(shù)成立,即n3+5n(nN+)能被

12、6整除.反思用數(shù)學(xué)歸納法證明有關(guān)整除性問(wèn)題的關(guān)鍵是尋找f(k+1)與f(k)之間的遞推關(guān)系,基本策略就是“往后退”,從f(k+1)中將f(k)分離出來(lái).目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三【變式訓(xùn)練2】

13、 用數(shù)學(xué)歸納法證明:1-(3+x)n(nN+)能被x+2整除.證明:(1)當(dāng)n=1時(shí),1-(3+x)=-(x+2),能被x+2整除,命題成立.(2)假設(shè)當(dāng)n=k(kN+,且k1)時(shí),1-(3+x)n能被x+2整除,則可設(shè)1-(3+x)k=(x+2)f(x) (f(x)為k-1次多項(xiàng)式).則當(dāng)n=k+1時(shí),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,即當(dāng)n=k+1時(shí)命題也成立.由(1)(2)可知,對(duì)nN+,1-(

14、3+x)n能被x+2整除.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三題型三 利用數(shù)學(xué)歸納法證明幾何問(wèn)題【例3】 平面內(nèi)有n個(gè)圓,任意兩個(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

15、)=n2-n+2(nN+)個(gè)部分.分析:因?yàn)閒(n)為n個(gè)圓把平面分割成的區(qū)域數(shù),如果再有一個(gè)圓和這n個(gè)圓相交,那么就有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(lái)的平面區(qū)域一分為二,因此,增加一個(gè)圓后,平面分成的區(qū)域數(shù)增加2n個(gè),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SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYAN

16、LIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三證明:(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,所以當(dāng)n=1時(shí)命題成立.(2)假設(shè)當(dāng)n=k(kN+,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),從(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)交點(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)

17、2-(k+1)+2.所以當(dāng)n=k+1時(shí),命題也成立.綜合(1)(2)可知,對(duì)一切nN+命題成立.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三反思對(duì)于幾何問(wèn)題的證明,可以從有限情形中歸納出一個(gè)變化的過(guò)程,或者說(shuō)體

18、會(huì)出是怎樣變化的,然后再去證明,也可以用“遞推”的辦法.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理題型一題型二題型三目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)

19、航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理12341用數(shù)學(xué)歸納法證明1+2+(2n+1)=(n+1)(2n+1)時(shí),在驗(yàn)證n=1成立時(shí),左邊所得的代數(shù)式是()A.1B.1+3C.1+2+3D.1+2+3+4答案:C目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANL

20、ITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1234答案:B 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHI

21、SHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理12343用數(shù)學(xué)歸納法證明關(guān)于n的恒等式,當(dāng)n=k時(shí),表達(dá)式為14+27+k(3k+1)=k(k+1)2,則當(dāng)n=k+1時(shí),表達(dá)式為.答案:14+27+k(3k+1)+(k+1)(3k+4)=(k+1)(k+2)2目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識(shí)梳理1234