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1、数学归纳法海选专题证明一些等式和不等式注意：（1）初始值（2）由 n=k 到 n=k+1 时注意增加的项数一定要用n=k 正确这个结论（3）特殊猜想证明（4）)1(11)1(12nnnnn一 用数学归纳法证明等式：1.用数学归纳法证明nnnnn2121112112141312112、用数学归纳法证明：123234 n(n 1)(n2)n n1 n2 n34(nN*)二用数学归纳法证明下述不等式:1.用数学归纳法证明11112()23n nNn2.设*,0Nnx且2n，求证：nxxn11.三整除问题：1.试证当 n 为自然数时，f(n)32n28n9 能被 64 整除2.若 5n23n1 1(n
2、N*)能被正整数m 整除，请写出m 的最大值，并给予证明解：当 n1 时，5123018，m8，(2 分)下证 5n23n11(nN*)能被 8 整除(3 分)当 n1 时已证；(4 分)假设当 nk(kN*)时命题成立，即5k23k11 能被 8整除(5 分)则当 nk 1 时，5k123k 15 5k6 3k11(6 分)(5k23k11)4(5k3k1)，(7 分)5k23k11 能被 8整除，而5k3k1为偶数，4(5k3k 1)也能被 8 整除，即当nk1 时命题也成立(8 分)3.用数学归纳法证明：*21Nnxn能被1x整除.四与数列有关问题：1已知正项数列na中，对于一切的*nN
3、均有21nnnaaa成立。（1）证明：数列na中的任意一项都小于1；（2）探究na与1n的大小，并证明你的结论.1解：（1）由21nnnaaa得21nnnaaa在数列na中0na，10na，20,01nnnaaa故数列na中的任意一项都小于1.（2）由（1）知1011na，那么2221111111()2442aaaa，由此猜想：1nan（n2）.下面用数学归纳法证明：当 n=2 时，显然成立；当 n=k 时（k2,k N）时，假设猜想正确，即112kak，那么22212221111111111()()242411kkkkkkaaaakkkkkk，当 n=k+1 时，猜想也正确综上所述，对于一切
4、*nN，都有1nan。2.已知数列na的前n项和为nS，通项公式为1nan，2211()2nnnSnf nSSn，（1）计算(1),(2),(3)fff的值；（2）比较()f n与 1 的大小，并用数学归纳法证明你的结论.解：（1）由已知213(1)122fS，4111113(2)23412fSS，62111119(3)345620fSS；（2）由（）知(1)1,(2)1ff；下面用数学归纳法证明：当3n时，()1f n（）由（）当3n时，()1f n；（）假设(3)nk k时，()1f k，即111()112f kkkk，那么11111(1)1222122f kkkkkk1111111122
5、2122kkkkkkk11111212222kkkk2(21)2(22)12(21)2(22)kkkkkkkk11112(21)(22)kkkk，所以当1nk时，()1f n也成立由（1）和（2）知，当3n时，()1f n所以当1n，和2n时，()1f n；当3n时，()1f n文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9
6、Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I
7、10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码
8、:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1
9、Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9
10、 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2
11、H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2
12、 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I103.已知数列na中，112a，21112nnnaaa(*)nN（1）求证：311 3(,)8 2a；（2）求证：当3n时，1|2|2nna解：（1）因为112a，所以22211111331(1)(1,)2222aaaa2分故22322211311 31(1)(,)2228 2aaaa4 分（2）当3n时，31132(2,2)82a，又111 312,288 28，所以311288a，即31|2|8a6 分假设当(3)nk k时，1|2|2kka则当1nk时，11|2|2|22|2kk
13、kaaa8 分111|222|222kk112k10 分即1nk时结论成立综上所述，当3n时，1|2|2nna4、已知正项数列na中，111,1()1nnnaaanNa。用数学归纳法证明：1()nnaanN。答案要点：当1n时，1213112aaa，12aa，所以，1n时，不等式成立；假设nk（*kN）时，1kkaa成立，则当1nk时，112111111(1)111kkkkkkkkkaaaaaaaaa11111kkaa110(1)(1)kkkkaaaa，文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2
14、ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z
15、10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I1
16、0文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:
17、CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y
18、5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9
19、HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H
20、5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10所以，1nk时，不等式成立综上所述，不等式1()nnaanN成立5、已知数列na满足11a，且11429nnnnaa aa（nN）（1）求123,a aa的值（2）由（1）猜想na的通项公式，并给出证明。解：（1）由11429nnnnaa aa得1921244nnnnaaaa，求得23471319,357aa
21、a 3 分（2）猜想6521nnan 5分证明：当n=1 时，猜想成立。6 分设当 n=k 时()kN时，猜想成立，即6521kkak，7 分则当 n=k+1 时，有111616(1)522654212(1)1421kkkkakakkk，所以当 n=k+1 时猜想也成立 9 分综合，猜想对任何nN都成立。10 分6、已知数列na中，an=n（n+1）（n+2）.又 Sn=kn（n+1）（n+2）（n+3），试确定常数k，使 S n恰为na的前 n 项的和，并用数学归纳法证明你的结论.解：由 a1=S1,k=14.下面用数学归纳法进行证明.1.当 n=1 时，命题显然成立；2.假设当 n=k(k
22、N*)时，命题成立,即 123+234+k（k+1）（k+2）=14k（k+1）（k+2）（k+3）,则 n=k+1 时，123+234+k（k+1）（k+2）+（k+1）（k+2）（k+3）=14k（k+1）（k+2）（k+3）+（k+1）（k+2）（k+3）=14(k+1)（k+1+1）（k+1+2）（k+1+3）即命题对n=k+1.成立由 1,2，命题对任意的正整数n 成立.7 已 知 正 项 数 列na中，111,1()1nnnaaanNa。用 数 学 归 纳 法 证 明：文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X
23、9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P
24、2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O
25、2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J
26、9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3
27、I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编
28、码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W
29、1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I101()nnaanN。8.已知数列na的各项都是正数且满足)4(21,110nnnaaaaNn（1）求21,aa（2）证明：21nnaa9、已知数列na满足*1121,232nnnaanaaN（）计算234,a a a；（）猜想数列的通项na，并利用数学归纳法证明解：（）由递推公式，
30、得121122321234232aaa，3457,68aa 3 分（）猜想：212nnan5 分证明：1n时，由已知，等式成立6 分设*()nk kN时，等式成立即212kkak7 分所以12122214212(1)122123426222(1)232kkkkakkkkkakakkkkk，所以1nk时，等式成立9 分根据可知，对任意*nN，等式成立即通项212nnan 10 分五与几何知识有关问题：1.空间内有*Nnn个不重合的平面，设这n个平面最多将空间分成na*Nn个部分.（1）求;,4321aaaa（2）写出na 关于*Nnn的表达式，并用数学归纳法证明.2.平面上有n 个圆，每两个圆交
31、于两点，每三个圆不过同一点，求证这n 个圆分平面为n2n2 个部分文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X
32、9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P
33、2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O
34、2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J
35、9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3
36、I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编
37、码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I103.n 个半圆的圆心在同一条直线l 上，这 n 个半圆每两个都相交，且都在直线l 的同侧，问这些半圆被
38、所有的交点最多分成多少段圆弧？分析：设这些半圆最多互相分成f(n)段圆弧，采用由特殊到一般的方法，进行猜想和论证当 n=2 时，由图(1)两个半圆交于一点，则分成4 段圆弧，故f(2)=4=22当 n=3 时，由图(2)三个半径交于三点，则分成9 段圆弧，故f(3)=9=32由 n=4 时，由图(3)三个半圆交于6 点，则分成16 段圆弧，故 f(4)=16=42由此猜想满足条件的n 个半圆互相分成圆弧段有f(n)=n2用数学归纳法证明如下：当 n=2 时，上面已证设 n=k 时，f(k)=k2，那么当 n=k+1 时，第 k+1 个半圆与原k 个半圆均相交，为获得最多圆弧，任意三个半圆不能交
39、于一点，所以第k+1 个半圆把原k 个半圆中的每一个半圆中的一段弧分成两段弧，这样就多出 k条圆弧；另外原k 个半圆把第k+1 个半圆分成k+1 段，这样又多出了k+1 段圆弧 f(k+1)=k2+k+(k+1)=k2+2k+1=(k+1)2 满足条件的k+1 个半圆被所有的交点最多分成(k+1)2段圆弧由、可知，满足条件的n 个半圆被所有的交点最多分成n2段圆弧说明：这里要注意；增加一个半圆时，圆弧段增加了多少条？可以从f(2)=4，f(3)=f(2)+2+3，f(4)=f(3)+3+4 中发现规律：f(k+1)=f(k)+k+(k+1)4.用,a b c d四个不同字母组成一个含1n*)(
40、Nn个字母的字符串，要求由a开始，相邻两个字母不同.例如1n时，排出的字符串是,ab ac ad；2n时排出的字符串是,aba abc abd aca acb acd ada adb adc，如图所示.记这含1n个字母的所有字符串中，排在最后一个的字母仍是a的字符串的种数为na.试用数学归纳法证明：*33(1)(N,1)4nnnann；a b c d n=1a b c d n=2a c d a b d a b c 文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP
41、1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D
42、8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB
43、8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W
44、1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ
45、8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10
46、D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文
47、档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10解（1）：证明：（）当1n时，因为10a，33(1)04，所以等式正确.（）假设nk时，等式正确，即*33(1)(N,1)4kkkakk，那么，1nk时，因为11133(1)4 333(1)33(1)33444kkkkkkkkkkkaa，这说明1nk时等式仍正确.据（），（）可知，*33(1)(N,1)4nnnann正确.-10分文档编码:CP1W
48、1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X
49、9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P
50、2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O2 ZJ8J9Z10D3I10文档编码:CP1W1Y5D8X9 HB8P2H5W1O