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2021年高二数学归纳法证明不等式.pdf

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1、学习必备欢迎下载第四讲：数学归纳法证明不等式数学归纳法证明不等式是高中选修的重点内容之一，包含数学归纳法的定义和数学归纳法证明基本步骤，用数学归纳法证明不等式。数学归纳法是高考考查的重点内容之一，在数列推理能力的考查中占有重要的地位。本讲主要复习数学归纳法的定义、数学归纳法证明基本步骤、用数学归纳法证明不等式的方法：作差比较法、作商比较法、综合法、分析法和放缩法，以及类比与猜想、抽象与概括、从特殊到一般等数学思想方法。在用数学归纳法证明不等式的具体过程中，要注意以下几点：（1）在从 n=k 到 n=k+1 的过程中，应分析清楚不等式两端（一般是左端）项数的变化，也就是要认清不等式的结构特征；（

2、2）瞄准当n=k+1 时的递推目标，有目的地进行放缩、分析；（3）活用起点的位置；（4）有的试题需要先作等价变换。例题精讲例 1、用数学归纳法证明nnnnn212111211214131211分析：该命题意图：本题主要考查数学归纳法定义，证明基本步骤证明：1 当 n=1 时，左边=1-21=21，右边=111=21，所以等式成立。2 假设当 n=k 时，等式成立，即kkkkk212111211214131211。那么，当n=k+1 时，221121211214131211kkkk221121212111kkkkk)22111(1212131214131211kkkkkk)1(211212131

3、21kkkkk这就是说，当n=k+1 时等式也成立。综上所述，等式对任何自然数n都成立。点评：数学归纳法是用于证明某些与自然数有关的命题的一种方法设要证命题为P（n）（1）证明当 n 取第一个值n0时，结论正确，即验证P（n0）正确；（2）假设 n=k（kN 且 kn0）时结论正确，证明当 n=k+1 时，结论也正确，即由 P（k）正确推出P（k+1）正确，根据（1），|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 1 页，共 16 页学习必备欢迎下载（2），就可以判定命题P（n）对于从n0开始的所有自然数n 都正确要证明的等式左边共2n 项，而

4、右边共n 项。f(k)与 f(k+1)相比较，左边增加两项，右边增加一项，并且二者右边的首项也不一样，因此在证明中采取了将11k与221k合并的变形方式，这是在分析了f(k)与 f(k+1)的差异和联系之后找到的方法。练习：1.用数学归纳法证明3kn3(n 3,n N)第一步应验证()A.n=1 B.n=2 C.n=3 D.n=4 解析：由题意知n3，应验证n=3.答案：C 2.用数学归纳法证明412n+3n+2能被 13 整除，其中nN 证明：(1)当 n=1 时，421+1+31+2=91 能被 13 整除(2)假设当 n=k 时，42k+1+3k+2能被 13 整除，则当n=k+1 时，

5、42(k+1)+1+3k+3=42k+142+3k+2342k+13+42k+13=42k+113+3(42k+1+3k+2)42k+113 能被 13 整除，42k+1+3k+2能被 13 整除当 n=k+1 时也成立.由知，当nN*时，42n+1+3n+2能被 13 整除.例 2、求证：*1115,(2,)1236nnNnnn分析：该命题意图：本题主要考查应用数学归纳法证明不等式的方法和一般步骤。用数学归纳法证明，要完成两个步骤，这两个步骤是缺一不可的但从证题的难易来分析，证明第二步是难点和关键，要充分利用归纳假设，做好命题从n=k 到 n=k+1 的转化，这个转化要求在变化过程中结构不变

6、证明：（1）当 n=2 时，右边=1111534566，不等式成立（2）假设当*(2,)nk kkN时命题成立，即11151236kkk则当1nk时，|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 2 页，共 16 页文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q

7、4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:

8、CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J

9、1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10

10、 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U1

11、0Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J

12、2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V

13、6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4学习必备欢迎下载111111(1)1(1)2331323(1)1111111()123313233151111()6313233151111()633333315115(3).63316kkkkkkkkkkkkkkkkkkkkkkk所以则当1nk时，不等式也成立由（1），（2）可知，原不等式对一切*2,nnN均成立点评：本题在由nk到1nk时的推证过程中，（1）一定要注意分析清楚命题的结构特征，即由nk到1nk时不等式左端项数的增减情况；（2）应用了放缩技巧：111111113.313233333

14、333331kkkkkkkk例 3、已知，*1111,23nSnNn，用数学归纳法证明：*21(2,)2nnSnnN证明：（1）当 n=2 时，22111132111234122S，命题成立（2）假设当*(2,)nk kkN时命题成立，即2111112322kkkS则当1nk时，112111111123221222kkkkkS|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 3 页，共 16 页文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6

15、F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4

16、文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:C

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19、HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10

20、Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2

21、 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4学习必备欢迎下载111111111111122122222221111211.22222kkkkkkkkkkkkk所以则当1nk时，不等式也成立由（1），（2）可知，原不等式对一切*2,nnN均成立点评：本题在由nk到1nk时的推证过程中，（1）不等式左端增加了2k项，而不是只增加了“112k”这一项，否则证题思路必然受阻

22、；（2）应用了放缩技巧：11111111111112.2122222222kkkkkkkk练习：1、证明不等式：分析1、数学归纳法的基本步骤：设 P(n)是关于自然数n的命题，若1 P(n0)成立(奠基)2 假设 P(k)成立(k n0)，可以推出P(k+1)成立(归纳)，则 P(n)对一切大于等于n0的自然数n 都成立.2、用数学归纳法证明不等式是较困难的课题，除运用证明不等式的几种基本方法外，经常使用的方法就是放缩法，针对目标，合理放缩，从而达到目标证明：（1）当 n=1 时，不等式成立（2）假设 n=k 时，不等式成立，即那么，|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*

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26、E8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8

27、T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档

28、编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2

29、K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4学习必备欢迎下载这就是说，n=k+1 时，不等式也成立根据（1）（2）可知不等式对n N+都成立2.求证：用数学

30、归纳法证明2*22()nnnN证明：（1）当 n=1 时，12221，不等式成立；当 n=2 时，22222，不等式成立；当 n=3 时，32223，不等式成立（2）假设当*(3,)nk kkN时不等式成立，即222kk则当1nk时，1222222(22)222(1)23kkkkkk，3k，223(3)(1)0kkkk，（*）从而122222(1)23(1)kkkkk，1222(1)kk即当1nk时，不等式也成立由（1），（2）可知，222nn对一切*nN都成立点评：因为在（*）处，当3k时才成立，故起点只证n=1 还不够，因此我们需注意命题的递推关系式中起点位置的推移3求证：23mem，其中

31、1m，且mN分析：此题是20XX 年广东高考数学试卷第21 题的适当变形，有两种证法证法一：用数学归纳法证明（1）当 m=2 时，44232e，不等式成立（2）假设*(2,)mk kkN时，有23kek，|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 5 页，共 16 页文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4

32、P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD

33、2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6

34、T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 Z

35、E8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8

36、T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档

37、编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2

38、K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4学习必备欢迎下载则2(1)22236kkeeek ek，2k，63(1)330kkk，即63(1)kk从而2(1)63(1)kekk，即1mk时，亦有23mem由（1）和（2）知，对1,mmN都成立证法二：作差、放缩，然后利用二项展开式和放缩法证明220122223(1 1)332(21)123(1211)21230mmmmmemmCCCmmmmmmmmmm当1m，且mN时，23mem例 4、（20XX 年江西省高考理科数学第21 题第（1）小

39、题，本小题满分12 分）已知数列na,:的各项都是正数且满足.),4(,21,110Nnaaaannn证明;,21Nnaann求数列na的通项公式an.分析：近年来高考对于数学归纳法的考查，加强了数列推理能力的考查。对数列进行了考查，和数学归纳法一起，成为压轴题。解：（1）方法一用数学归纳法证明：1当 n=1 时，,23)4(21,10010aaaa210aa，命题正确.2假设 n=k 时有.21kkaa则111111,(4)(4)22kkkkkknkaaaaaa时11111112()()()()(4).22kkkkkkkkkkaaaaaaaaaa而1110,40,0.kkkkkkaaaaaa

40、|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 6 页，共 16 页文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 Z

41、E8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8

42、T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档

43、编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2

44、K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4

45、P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD

46、2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4学习必备欢迎下载又2111(4)4(2)2.22kk

47、kkaaaa1kn时命题正确.由 1、2知，对一切nN 时有.21nnaa方法二：用数学归纳法证明：1当 n=1 时，,23)4(21,10010aaaa2010aa；2假设 n=k 时有21kkaa成立，令)4(21)(xxxf，)(xf在0，2上单调递增，所以由假设有：),2()()(1fafafkk即),24(221)4(21)4(2111kkkkaaaa也即当 n=k+1 时21kkaa成立，所以对一切2,1kkaaNn有（2）下面来求数列的通项：,4)2(21)4(2121nnnnaaaa所以21)2()2(2nnaa2,nnba令则21222221 222121111111()()

48、()222222nnnnnnnbbbbb又 bn=1，所以211(),2nnb21122()2nnnab即点评：本题问给出的两种方法均是用数学归纳法证明，所不同的是：方法一采用了作差比较法；方法二利用了函数的单调性|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 7 页，共 16 页文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K

49、2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P

50、10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2U10Q6T9J2 ZE8V6F8T2Q4文档编码:CJ2K2J1K4P10 HD2

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