例15.(2008年厦门市质检) 已知函数f(x)是在(0,??)上处处可导的函数,若x?f'(x)? (I)求证：函数g(x)?f(x)上是增函数； 在(0,??)xf(x)在x?0上恒成立.
(II)当x1?0,x2?0时,证明:f(x1)?f(x2)?f(x1?x2)；
(III)已知不等式ln(1?x)?x在x??1且x?0时恒成立， 求证：
1111nln22?2ln32?2ln42???ln(n?1)2?222(n?1)(n?2)234(n?1)(n?N*).
解析:(I)g'(x)? (II)因为
,所以函数上是增函数 f'(x)x?f(x)f(x)?0g(x)?在(0,??)2xxg(x)?f(x)在(0,??)x上是增函数,所以
f(x1)f(x1?x2)x1 ??f(x1)??f(x1?x2)x1x1?x2x1?x2f(x2)f(x1?x2)x2 ??f(x2)??f(x1?x2)x2x1?x2x1?x2 两式相加后可以得到f(x1)?f(x2)?f(x1?x2) (3)
f(x1)f(x1?x2???xn)x1??f(x1)??f(x1?x2???xn)x1x1?x2???xnx1?x2???xnf(x2)f(x1?x2???xn)x2…… ??f(x2)??f(x1?x2???xn)x2x1?x2???xnx1?x2???xnf(xn)f(x1?x2???xn)xn ??f(xn)??f(x1?x2???xn)xnx1?x2???xnx1?x2???xn
相加后可以得到: f(x1)?f(x2)???f(xn)?f(x1?x2???xn)
所以x1lnx1?x2lnx2?x3lnx3???xnlnxn?(x1?x2???xn)ln(x1?x2???xn) 令
xn?1(1?n)2,有
?1111?1111??111? 2222????22ln2?32ln3?42ln4???(n?1)2ln(n?1)?????22?32?42???(n?1)2???ln??22?32???(n?1)2????????1??11?n?111??111?????????????????????ln??????22322(n?1)(n?2)(n?1)n?(n?1)2???n?1??2n?2????2?13?2
所以
1111nln22?2ln32?2ln42???ln(n?1)2?222(n?1)(n?2)234(n?1)?ln(n?1)2ln41? ?1??ln4???(n?1)(n?2)(n?1)(n?2)?n?1n?2?(n?N*).
(方法二)ln(n?1)2(n?1)2 所以
11111?nln4?1ln22?2ln32?2ln42???ln(n?1)2?ln4????22234(n?1)?2n?2?2(n?2)
又ln4?1?1,所以1 111nln22?2ln32?2ln42???ln(n?1)2?(n?N*).22n?12(n?1)(n?2)234(n?1)6 / 26
三、分式放缩
姐妹不等式:b?b?m(b?a?0,m?0)和b?b?m(a?b?0,m?0)
aa?maa?m 记忆口诀”小者小,大者大”
解释:看b,若b小,则不等号是小于号,反之. 例19. 姐妹不等式:(1?1)(1?1)(1?1)?(1?1)?2n?1和
352n?11111(1?)(1?)(1?)?(1?)?2462n12n?112n?1也可以表示成为
和1?3?5???(2n?1)?2?4?6??2n?2n?12?4?6???2n1?3?5???(2n?1)a解析: 利用假分数的一个性质b?b?m(b?a?0,m?0)可得
a?m 2?4?6?2n?1352n?1?(2?4?6?3572n?11352n?1 ????????(2n?1)2462n2462n1351112n2)?2n?1. )?2n?1即(1?1)(1?)(1?)?(1?352n?12n?1
111例20.证明:(1?1)(1?)(1?)?(1?)?33n?1.
473n?2解析: 运用两次次分式放缩:
2583n?13693n ??????.?????1473n?22583n?1 (加1) (加2)
2583n?147103n?1 ??????.?????1473n?23693n 相乘,可以得到:
3n?1?47103n?11473n?2?258 ???????(3n?1)????????.?????3n?2?2583n?12583n?1?1472 所以有(1?1)(1?1)(1?1)?(1?1)?33n?1.
473n?2
四、分类放缩
例21.求证:1?1?1??23?1n? 2?12n 解析: 1?1?1???2311111111?1??(?)?(3?3?3?3)???
2442?12222n(1111n1n?n???n)?n??(1?n)? n2222222n例22.(2004年全国高中数学联赛加试改编) 在平面直角坐标系xoy中, y轴正半轴上的点列?A?与曲线
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y?2x（x≥0）上的点列?Bn?满足OA?OB?1，直线AnBn在x轴上的截距为an.点Bn的横坐标为bn,n?N?.
nnn(1)证明an>an?1>4，n?N?; (2)证明有n0?N?，使得对?n?n0都有b2?b3???bn?bn?1 b1b2bn?1bn 解析:(1) 依题设有：A bn2?2bn?n?1?，由OB?1得： n?0,?,Bnbn,2bn,?bn?0?n?n???,又直线AnBn在11,?bn??1?1,n?N*22nn an?bnx轴上的截距为an满足 Q2n2bn?an?0??? 1??1?2bn????0???bn?0?n??n??1?n2bn?1?n2bn2?0,bn?2?1nbn2 ?an?bn1?n2bnbn12??2??bn?2?2bn?4?an?12?1?1?2?212?1 2nn1?2nbnnbnnbn1?n2bn1，有an?an?1?4,n?N* ?0n?1??显然，对于1?n (2)证明：设 1?1?n2cn?1?，则 bn?1,n?N*bn?1?11??n2?2???n?n?1?2???1?1?1 n211?1??122n?n?1?1cn??n?1?21?1?1n2??1?1?1??22n?1n2n?1112n?1??????222?n?1?21?1?n?1??221?1?2?n?1???n2n2??Q?2n?1??n?2??2?n?1??n?0,?cn?21,n?N* n?2设Sn?c1?c2?L?cn,n?N*，则当n?2k?2?1?k?N*?时， 1111?11??11??11Sn???L?k?k??????2?L?3???k?1?L?k342?122??2?12?34??2?1?2?111k?1。 ?22?3?L?2k?1?k?22222? ??所以，取n0?24009?2，对?n?n0都有： ?bn?1??b2??b3?4017?1??????1??1????1??S?S??2008 nn?b??b?0??bn?21?2????故有b2?b3???bn?bn?1 b1b2bn?1bn例23.(2007年泉州市高三质检) 已知函数f(x)?x2?bx?c(b?1,c?R)，若f(x)的定义域为[－1，0]，值域也为[－1，0].若数列{bn}满足bn?f(n)(n?N*)，记数列{bn}的前n3n项和为Tn，问是否存在正常数A，使得对于 任意正整数n都有Tn?A？并证明你的结论。 解析:首先求出f(x)?x2?2x,∵bn?f(n)n2?2n1??n3n3n ∴Tn?b1?b2?b3???bn?1?1?1???1,∵1?1?2?1?1,1?1?1?1?4?1?1,… 23n82344256788 / 26 1111,故当n?2k时,kk?1????2??T??1, n2k?1?12k?1?22k2k22?1因此，对任何常数A，设m是不小于A的最小正整数， 则当n?22m?2时,必有T?2m?2?1?m?A. n2故不存在常数A使Tn?A对所有n?2的正整数恒成立. ?x?0,例24.(2008年中学教学参考)设不等式组?y?0,表示的平面区域为D, ??y??nx?3n?n设D内整数坐标点的个数为an.设Sn?1?1???1, 当n?2时,求证:1?1?1???1?7n?11. nan?1an?2a2na1a2a3a2n36解析:容易得到an?3n,所以,要证1a1?1117n?11只要证Sn?????2a2a3a2n36?1?1117n?11,因为 ????n?23212S2n?1?1377n?11,所以原命题得1111111111?(?)?(???)???(n?1?n?1???n?1??T21?T22???T2n?1??(n?1)?22121223456782?12?22证 五、迭代放缩 例25. 已知xn?1?xn?4n,x1?1,求证:当n?2时,?|xi?2|?2?21?n xn?1i?1 解析:通过迭代的方法得到xn?2?1,然后相加就可以得到结论 n?12例26. 设Snsin1!sin2!sinn!,求证:对任意的正整数?1?2???n222|Sn?k?Sn|?|sin(n?1)!sin(n?2)!sin(n?k) ????|n?1n?2222n?k1 k,若k≥n恒有:|Sn+k－Sn|< n 解析: ?|sin(n?1)!sin(n?2)!sin(n?k)111|?||???||?n?1?n?2???n?kn?1n?2n?k2222221111111(?2???k)?n?(1?k)?nn2222222 ? 所以|S11 ?2nn 又2n01n?(1?1)n?Cn?Cn???Cn?n n?k?Sn|? 六、借助数列递推关系 例27.求证:1?1?3?1?3?5??22?42?4?6?1?3?5???(2n?1) ?2n?2?12?4?6???2n 解析: 设aan?1?n?1?3?5???(2n?1)则 2?4?6???2n2n?1an?2(n?1)an?1?2nan?an,从而 2(n?1)an?2(n?1)an?1?2nan,相加后就可以得到 a1?a2???an?2(n?1)an?1?2a1?2(n?1)?12n?3?1?(2n?2)?12n?2?1 9 / 26 所以1?1?3?1?3?5??22?42?4?6?1?3?5???(2n?1)?2n?2?1 2?4?6???2n 例28. 求证:1?1?3?1?3?5??22?42?4?6?1?3?5???(2n?1)?2n?1?1 2?4?6???2n 解析: 设a?1?3?5???(2n?1)则 n2?4?6???2nan?1?2n?1,从而 an?[2(n?1)?1]an?1?(2n?1)an?an?12(n?1)an?1?[2(n?1)?1]an?1?(2n?1)an,相加后就可以得到 a1?a2???an?(2n?1)an?1?3a1?(2n?1)?12n?1? 3?2n?1?12 例29. 若a1?1,an?1?an?n?1,求证:1?a111????2(n?1?1)a2an 解析: an?2?an?1?n?2?an?an?1?1?1an?1?an?2?an 所以就有1a1? 111?????an?1?an?a2?a1?2an?1an?a2?2n?1?2a2ana1 七、分类讨论 例30.已知数列{an}的前n项和Sn满足Sn?2an?(?1)n,n?1.证明：对任意的整数m?4，有 解析:容易得到a2n?2, 2?(?1)n?1.31117 ?????a4a5am8n??? 由于通项中含有(?1)，很难直接放缩，考虑分项讨论： 当n?3且n为奇数时 ?n 1131132n?2?2n?1??(n?2?n?1)??2n?3anan?122?12?122?2n?1?2n?2?132n?2?2n?1311（减项放缩），于是 ???(n?2?n?1)2n?322222 ①当m?4且m为偶数时1?1???1?1?(1?1)???(1?1) a4a5a6a4a5am131111311137??(3?4???m?2)????(1?m?4)???. 222224288222am?1am②当m?4且m为奇数时①②得证。 八、线性规划型放缩 （添项放缩）由①知11111117由?????1?1???1?1??????.a4a5ama4a5a4a5amam?18amam?1 例31. 设函数f(x)?2x?1.若对一切x?R，?3?af(x)?b?3，求a?b的最大值。 x2?2 解析:由 1?(x?2)2(x?1)2(f(x)?)(f(1)?1)?22(x2?2)2知(f(x)?1)(f(1)?1)?0 即 2 1??f(x)?12由此再由f(x)的单调性可以知道f(x)的最小值为?1，最大值为1 210 / 26
高考数学-压轴题-放缩法技巧全总结(最强大)
