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

《高中數(shù)學 第二章 幾個重要的不等式 2.3 數(shù)學歸納法與貝努利不等式 2.3.1 數(shù)學歸納法課件 北師大版選修45》由會員分享，可在線閱讀，更多相關《高中數(shù)學 第二章 幾個重要的不等式 2.3 數(shù)學歸納法與貝努利不等式 2.3.1 數(shù)學歸納法課件 北師大版選修45（21頁珍藏版）》請在裝配圖網(wǎng)上搜索。

1、3數(shù)學歸納法與貝努利不等式3 3.1 1數(shù)學歸納法目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1.理解數(shù)學歸納法的原理和實質(zhì).2.掌握用數(shù)學歸納法證明與正整數(shù)有關的命題的兩個步驟,并能靈活運用.目標導航DIANLITOUXI典例

2、透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理對數(shù)學歸納法的理解(1)數(shù)學歸納法原理:數(shù)學歸納法原理是設有一個關于正整數(shù)n的命題,若當n取第1個值n0時該命題成立,又在假設當n取第k個值時該命題成立后可以推出n取第k+1個值時該命題成立,則該命題對一切自然數(shù)nn0

3、都成立.(2)數(shù)學歸納法:數(shù)學歸納法可以用于證明與正整數(shù)有關的命題.證明需要經(jīng)過兩個步驟:驗證當n取第一個值n0(如n0=1或2等)時命題正確.假設當n=k時(kN+,kn0)命題正確,證明當n=k+1時命題也正確.在完成了上述兩個步驟之后,就可以斷定命題對于從n0開始的所有正整數(shù)都正確.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳

4、理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理【做一做1】 在用數(shù)學歸納法證明多邊形內(nèi)角和定理時,第一步應檢驗()A.當n=1時成立B.當n=2時成立C.當n=3時成立D.當n=4時成立解析:多邊形中至少有三條邊,故應先驗證當n=3時成立.答案:C目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISH

5、ISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理A.當n=k+1時等式成立B.當n=k+2時等式成立C.當n=2k+2時等式成立D.當n=2(k+2)時等式成立解析:因為已假設當n=k(k2,且k為偶數(shù))時命題為真,即當n=k+2時命題為真.而選項中n=k+1為奇數(shù),n=2k+2和n=2(k+2)均不滿足遞推關系,所以只有n=k+2滿足條件.答案:B目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨

6、堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理A.2kB.2k-1C.2k-1D.2k+1答案:A目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI

7、知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型一 用數(shù)學歸納法證明恒等問題 分析:在證明時,要嚴格按數(shù)學歸納法的步驟進行,并要特別注意當n=k+1時等式兩邊的式子,與當n=k時等式兩邊的式子之間的聯(lián)系,明確增加了哪些項,減少了哪些項.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂

8、演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三反思在解本題時,當由n=k到n=k+1時,等式的左邊增加了一項,這里容易因忽略而出錯.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUIT

9、ANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三【變式訓練1】 用數(shù)學歸納法證明: 目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型二 用數(shù)學歸納法證明整除問題【例2】用數(shù)學歸納法證

10、明:n3+5n(nN+)能被6整除.分析:這是一個與整除有關的命題,它涉及全體正整數(shù),第一步應證明當n=1時成立,第二步應明確目標,在假設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.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練Z

11、HISHISHULI知識梳理題型一題型二題型三證明:(1)當n=1時,n3+5n=6顯然能被6整除,命題成立.(2)假設當n=k(kN+,且k1)時,命題成立,即k3+5k能被6整除.則當n=k+1時,(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n=k+1時,命題也成立.由(1)(2)知,命題對一切正整數(shù)成立,即n3+5n(nN+)能被

12、6整除.反思用數(shù)學歸納法證明有關整除性問題的關鍵是尋找f(k+1)與f(k)之間的遞推關系,基本策略就是“往后退”,從f(k+1)中將f(k)分離出來.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三【變式訓練2】

13、 用數(shù)學歸納法證明:1-(3+x)n(nN+)能被x+2整除.證明:(1)當n=1時,1-(3+x)=-(x+2),能被x+2整除,命題成立.(2)假設當n=k(kN+,且k1)時,1-(3+x)n能被x+2整除,則可設1-(3+x)k=(x+2)f(x) (f(x)為k-1次多項式).則當n=k+1時,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n=k+1時命題也成立.由(1)(2)可知,對nN+,1-(

14、3+x)n能被x+2整除.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型三 利用數(shù)學歸納法證明幾何問題【例3】 平面內(nèi)有n個圓,任意兩個圓都相交于兩點,任意三個圓不相交于同一點,求證:這n個圓將平面分成f(n

15、)=n2-n+2(nN+)個部分.分析:因為f(n)為n個圓把平面分割成的區(qū)域數(shù),如果再有一個圓和這n個圓相交,那么就有2n個交點,這些交點將增加的這個圓分成2n段弧,且每一段弧又將原來的平面區(qū)域一分為二,因此,增加一個圓后,平面分成的區(qū)域數(shù)增加2n個,即f(n+1)=f(n)+2n.有了上述關系,數(shù)學歸納法的第二步證明可迎刃而解.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYAN

16、LIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三證明:(1)當n=1時,一個圓將平面分成兩個部分,且f(1)=1-1+2=2,所以當n=1時命題成立.(2)假設當n=k(kN+,k1)時命題成立,即k個圓把平面分成f(k)=k2-k+2個部分,則當n=k+1時,從(k+1)個圓中任取一個圓O,剩下的k個圓將平面分成f(k)個部分,而圓O與k個圓有2k個交點,這2k個交點將圓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n=k+1時,命題也成立.綜合(1)(2)可知,對一切nN+命題成立.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三反思對于幾何問題的證明,可以從有限情形中歸納出一個變化的過程,或者說體

18、會出是怎樣變化的,然后再去證明,也可以用“遞推”的辦法.目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導

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

20、ITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1234答案:B 目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHI

21、SHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理12343用數(shù)學歸納法證明關于n的恒等式,當n=k時,表達式為14+27+k(3k+1)=k(k+1)2,則當n=k+1時,表達式為.答案:14+27+k(3k+1)+(k+1)(3k+4)=(k+1)(k+2)2目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1234