2022年2019届高考数学大一轮总复习冲刺第十三章推理与证明算法复数13.3数学归纳法及其应用精品学案理北师大版 .pdf

《2022年2019届高考数学大一轮总复习冲刺第十三章推理与证明算法复数13.3数学归纳法及其应用精品学案理北师大版 .pdf》由会员分享，可在线阅读，更多相关《2022年2019届高考数学大一轮总复习冲刺第十三章推理与证明算法复数13.3数学归纳法及其应用精品学案理北师大版 .pdf（17页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、,最新教学推荐,1 13.3数学归纳法最新考纲考情考向分析1.了解数学归纳法的原理2.能用数学归纳法证明一些简单的数学命题.以了解数学归纳法的原理为主，会用数学归纳法证明与数列有关或与不等式有关的等式或不等式在高考中以解答题形式出现，属高档题.数学归纳法数学归纳法是用来证明某些与正整数n有关的数学命题的一种方法它的基本步骤是：(1)验证：当n取第一个值n0(如n01 或 2 等)时，命题成立；(2)在假设当nk(kN，kn0)时命题成立的前提下，推出当nk 1 时，命题成立根据(1)(2)可以断定命题对一切从n0开始的正整数n都成立题组一思考辨析1判断下列结论是否正确(请在括号中打“”或“”)

2、(1)用数学归纳法证明问题时，第一步是验证当n1 时结论成立()(2)所有与正整数有关的数学命题都必须用数学归纳法证明()(3)用数学归纳法证明问题时，归纳假设可以不用()(4)不论是等式还是不等式，用数学归纳法证明时，由nk到nk1 时，项数都增加了一项()(5)用数学归纳法证明等式“1 222,2n22n31”，验证n1 时，左边式子应为122223.()(6)用数学归纳法证明凸n边形的内角和公式时，n03.()题组二教材改编2在应用数学归纳法证明凸n边形的对角线为12n(n3)条时，第一步检验n等于(),最新教学推荐,2 A1 B 2 C3 D 4 答案C 解析凸n边形边数最小时是三角形

3、，故第一步检验n3.3 已知 an 满足an 1a2nnan1，nN，且a12，则a2_，a3_，a4_，猜想an_.答案3 4 5 n1 题组三易错自纠4用数学归纳法证明1aa2,an11an21a(a1，n N)，在验证n1 时，等式左边的项是()A1 B 1aC1aa2D 1aa2a3答案C 解析当n1 时，n1 2，左边 1a1a21aa2.5对于不等式n2nn 1(nN)，某同学用数学归纳法证明的过程如下：(1)当n1 时，12111，不等式成立(2)假设当nk(kN)时，不等式成立，即k2kk1，则当nk 1时，k12k1k23k20，整数p1，nN.(1)证明：当x1 且x0 时

4、，(1 x)p1px；(2)数列 an满足a11pc，an1p1pancpa1pn.证明：anan11pc.证明(1)当p2 时，(1 x)21 2xx212x，原不等式成立假设当pk(k2，kN)时，不等式(1x)k1kx成立则当pk1 时，(1x)k1(1 x)(1 x)k(1 x)(1kx)1(k1)xkx21(k1)x.所以当pk1 时，原不等式也成立综合可得，当x1，且x0 时，对一切整数p1，不等式(1 x)p1px均成立(2)方法一当n1 时，由题设知a11pc成立假设当nk(k1，kN)时，不等式ak1pc成立由an1p1pancpa1pn易知an0，nN.则当nk1 时，ak

5、1akp1pcpapk11pcapk1.由ak1pc 0 得 11p1pcapk11p1pcapk1capk.文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:

6、CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T

7、2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:

8、CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T

9、2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:

10、CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T

11、2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5,最新教学推荐,5 因此apk1c，即ak11pc.所以当nk1 时，不等式an1pc也成立综合可得，对一切正整数n，不等式an1pc均成立再由an1an11pcapn1 可得an1an1，即an1an 11pc，nN.方法二设f(x)p1pxcpx

12、1p，x1pc，则xpc，并且f(x)p1pcp(1 p)xpp1p1cxp0，x1pc.由此可得，f(x)在 1pc，)上是增加的，因而，当x1pc时，f(x)f(1pc)1pc.当n 1时，由a11pc0，即1pac可知a2p1pa1cp11paa111pca1p11pc，从而a1a21pc.故当n 1时，不等式anan11pc成立假设当nk(k1，kN)时，不等式akak11pc成立，则当nk1 时，f(ak)f(ak1)f(1pc)，即有ak1ak 21pc.所以当nk1 时，原不等式也成立文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7

13、Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2

14、 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7

15、Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2

16、 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7

17、Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2

18、 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7

19、Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5,最新教学推荐,6 综合可得，对一切正整数n，不等式anan11pc均成立思维升华数学归纳法证明不等式的适用范围及关键(1)适用范围：当遇到与正整数n有关的不等式证明时，若用其他办法不容易证，则可考虑应用数学归纳法(2)关键：由nk时命题成立证nk1 时命题也成立，在归纳假设使用后可运用比较法、综合法、分析法、放缩法等来加以证明，充分应用基本不等式、不等式的性质等放缩技巧，使问题得以简化跟踪训练(2018衡水调研)若函数f(x)x22x3，定义数列 x

20、n如下：x12，xn1是过点P(4,5)，Qn(xn，f(xn)(nN)的直线PQn与x轴的交点的横坐标，试运用数学归纳法证明：2xnxn13.证明当n1 时，x12，f(x1)3，Q1(2，3)所以直线PQ1的方程为y4x11，令y0，得x2114，因此 2x1x23，即n1 时结论成立假设当nk(k1，kN)时，结论成立，即2xkxk 13.当nk1 时，直线PQk 1的方程为y5f xk15xk14(x4)又f(xk1)x2k12xk13，代入上式，令y0，得xk234xk12xk1452xk 1，由归纳假设，2xk13，xk2452xk10，即xk1xk2，所以 2xk1xk23，即当

21、nk1 时，结论成立由知对任意的正整数n,2xnxn11 时，对x(0，a1，有(x)0，(x)在(0，a 1 上是减少的，(a1)1 时，存在x0，使(x)0(nN)猜想an的通项公式，并用数学归纳法加以证明解分别令n1,2,3，得2a1a21 1，2a1a2a222，2a1a2a3a233，an0，a11，a2 2，a33，猜想：ann.由 2Sna2nn，可知，当n2 时，2Sn 1a2n 1(n1)，得2ana2na2n11，即a2n2ana2n1 1.()当n2 时，a222a2 12 1，a20，a22.()假设当nk(k2，kN)时，akk，那么当nk1 时，a2k12ak1a2

22、k12ak 1k21，即ak 1(k1)ak 1(k1)0，ak 10，k2，ak 1(k1)0，ak 1k1，即当nk1 时也成立ann(n2)，显然当n1 时，也成立，故对于一切nN，均有ann.命题点 3 存在性问题的证明典例设a11，an1a2n2an2b(nN)(1)若b1，求a2，a3及数列 an 的通项公式；(2)若b 1，问：是否存在实数c使得a2nca2n1对所有nN成立？证明你的结论解(1)方法一a22，a321.再由题设条件知(an11)2(an1)2 1.从而(an1)2是首项为0，公差为1 的等差数列，故(an1)2n1，即ann11(nN)方法二a22，a321.可

23、写为a1111，a221 1，a331 1.文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3

24、U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 H

25、E6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3

26、U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 H

27、E6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3

28、U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 H

29、E6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5,最新教学推荐,9 因此猜想ann1 1.下面用数学归纳法证明上式：当n1 时结论显然成立假设当nk(k1，kN)时结论成立，即akk1 1，则ak1ak1211k1 1 1 k1 11.所以当nk1 时结论成立所以ann 11(nN)(2)方法一设f(x)x 1211，则an1f(an)令cf(c)，即cc121

30、1，解得c14.下面用数学归纳法证明加强命题：a2nca2n11.当n1 时，a2f(1)0，a3f(a2)f(0)2 1，所以a214a31，结论成立假设当nk(k1，kN)时结论成立，即a2kca2k1f(a2k1)f(1)a2，即 1ca2k2a2.再由f(x)在(，1 上为减函数，得cf(c)f(a2k2)f(a2)a31，故ca2k31.因此a2(k 1)ca2(k1)11.这就是说，当nk1 时结论成立综上，符合条件的c存在，其中一个值为c14.方法二设f(x)x1211，则an1f(an)先证：0an1(nN)当n1 时，结论显然成立假设当nk(k1，kN)时结论成立，即0ak1

31、.易知f(x)在(，1 上为减函数，从而0f(1)f(ak)f(0)211，即 0ak11.文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2

32、T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1

33、M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2

34、T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1

35、M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2

36、T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1

37、M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5,最新教学推荐,10 这就是说，当nk1 时结论成立故成立再证：a2na2n1(nN)当n1 时，a2f(1)0，a3f(a2)f(0)2 1，有a2a3，即n1 时成立假设当nk(k1，kN)时，结论成立，即a2kf(a2k 1)a2k2，a2(k1)f(a2k1)

38、f(a2k2)a2(k 1)1.这就是说，当nk1 时成立，所以对一切nN成立由得a2na22n2a2n2 1，即(a2n1)2a22n2a2n 2，因此a2nf(a2n1)，即a2n1a2n2，所以a2n1a22n1 2a2n 121.解得a2n114.综上，由知存在c14使得a2nc0，an 10，ana2n0，文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L

39、5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文

40、档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L

41、5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文

42、档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L

43、5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文

44、档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5,最新教学推荐,11 0an1，故数列 a

45、n 中的任何一项都小于1.(2)由(1)知 0a1111，那么a2a1a21a1122141412，由此猜想an1n.下面用数学归纳法证明：当n2，且nN时猜想正确当n 2时已证；假设当nk(k2，且kN)时，有ak1k成立，那么1k12，ak1aka2kak122141k122141k1k2k1k2k1k211k1，当nk1 时，猜想正确综上所述，对于一切n N，都有an1n.归纳猜想证明问题典例 (12 分)数列 an满足Sn2nan(nN)(1)计算a1，a2，a3，a4，并由此猜想通项公式an；(2)证明(1)中的猜想思维点拨 (1)由S1a1算出a1；由anSnSn1算出a2，a3，

46、a4，观察所得数值的特征猜出通项公式(2)用数学归纳法证明规范解答(1)解当n1 时，a1S12a1，a1 1；当n2 时，a1a2S222a2，a232；当n3 时，a1a2a3S323a3，a374；当n4 时，a1a2a3a4S424a4，a4158.2分 由此猜想an2n12n1(nN)4 分 文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2

47、 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7

48、Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2

49、 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7

50、Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2 ZY7J1M8P3U5文档编码:CM7Y3Q2U2T4 HE6L5S4T2X2