解析:
?1?,从而an2?an?12?2,所以有 2?an??a??a?2k?1?n?1a?n?1??22 an2?(an2?an?12)?(an?12?an?22)???(a22?a12)?a12?2(n?1)?1?2n?1,所以an?2n?1 又a2n?1?,所以an2?an?12?3,所以有 2???a??a?3n?1k?1?an?1???2 an2?(an2?an?12)?(an?12?an?22)???(a22?a12)?a12?3(n?1)?1?3n?2所以an?3n?2 所以综上有2n?1?an?3n?2(n?2). 引申:已知数列{an}满足:a?1,a?a?1,求证: n11n?1n?aank?1?2n?1.
122n?1?2n?3k解析:由上可知an?2n?1,又
2n?1?2n?1?2n?32,所以1an?2n?1??2n?1?2n?3
n 从而?1?1?3?1?5?3???2n?1?2n?3?2n?1(n?2)
k?1akn 又当n?1时,1?1,所以综上有?1?2n?1.
a1k?1ak
同题引申: (2008年浙江高考试题)已知数列?a?,an?0,a1?0,an?12?an?1?1?an2(n?N?).
n记Sn?a1?a2???an,
T(1)an?an?1;
n?.求证:当n?N?时. 111????1?a1(1?a1)(1?a2)(1?a1)(1?a2)?(1?an) ★(3)Tn?3.
(2)Sn?n?2;
解析:(1)an?12?an2?1?an?1,猜想an?1,下面用数学归纳法证明: (i)当n?1时,a1?1,结论成立;
(ii)假设当n?k(k?1)时,ak?1,则n?k?1(k?1)时,ak?12?ak?1?1?ak2 从而ak?12?ak?1?2?an?1?1,所以0?ak?1?1 所以综上有0?an?1,故an?12?an2?0?an?1?an
(2)因为an?12?an2?1?an?1则a22?a12?1?a2,a32?a22?1?a3,…, an?12?an2?1?an?1,相加后可以得
到: an?12?a12?n?(a2?a3???an?1)?Sn?1?n?an?12,所以
Sn?n?1?an?n?2,所以Sn?n?2
2 (3)因为an?12?an?1?1?an2?2an,从而a
n?1?1?2an,有1a?n?1,所以有
an?11?an?12anaaaa1?n?1?n?3?n?n1?1,从而
(1?a3)?(1?an)(1?an?1)2an2an?12a22a2aa?1,所以 11?n?n1?1??n(1?a1)(1?a2)(1?a3)?(1?an)(1?an?1)2a21?a22n?1aa,所以 11?n21n??nn(1?a1)(1?a2)(1?a3)?(1?an)2a21?a22?2Tn?1?aaa111112?3?4???nn?1???2???n?2??1?1?3 2?21?a2221?a222225?1 所以综上有Tn?3.
例61.(2008年陕西省高考试题)已知数列{an}的首项 (1)证明:对任意的x?0,a≥1?n1?x2 (2)证明:a?a??a?n.
12na1?3,an?15?3an,n?1，2，． 2an?11?2?2，; ?x?,n?1，2?n(1?x)?3?n?1n 解析:(1)依题,容易得到an?32?3n?1?2,要证x?0,11?2?2，, an≥??x?,n?1，n2?n31?x(1?x)?3?即证1?2?1?n31?x1?2221??x?1?1???n?2?n2(1?x)?3(1?x)2?1?x3(1?x)
n即证2?2?3n1?x3(1?x)23n?2,设t?1所以即证明?(t)??2?3n?t2?2t?2?1?0(0?t?1) ?1?01?x3n3n3n2,这是显然成立的. ?1?0n3≥11?2?,n?1，2，??x?2?n1?x(1?x)?3?从而?(1)?0,即?2?3n?2?所以综上有对任意的x?0,a (法二)
n11?2?2? ??x??1?12??1?1?x?2?n?n1?x(1?x)?3?1?x(1?x)?3?221?1?1?1≤an??????a?(1?x)n??an???1?xa(1?x)2a1?xa?nn??n??11?1?x(1?x)2,?原不等式成立．
(2)由(1)知，对任意的x?0，有
a1?a2??an≥11?2?11?2???x????x??2?2?21?x(1?x)?3?1?x(1?x)?3??n1?22????1?x(1?x)2?3322?n3?11?2???x? 2?n1?x(1?x)?3??2?． ?nx?n3??取
1?22x???2?n?332?1?， ?1?n?11??3?3????1?n????n?1?1?n?3????3?则
a1?a2?nn2n2?an≥??1n?11?1?1??1?n?n?1?n3n?3?．
?原不等式成立．
十四、经典题目方法探究
探究1.(2008年福建省高考)已知函数f(x)?ln(1?x)?x.若f(x)在区间[0,n](n?N*)上的最小值为bn, 令an?ln(1?n)?bn.求证:a1?a1?a3???a1?a3?a5???a2n?1?2a?1?1.
na2a2?a4a2?a4?a6???a2n证明:首先:可以得到an?nn.先证明1?3?5???(2n?1)?2?4?6???2n41 2n?12n?12n?12 (方法一) ?1?3?5???(2n?1)??1?3?3?5???(2n?1)(2n?1)?1?1 所以1?3?5???(2n?1)?222???2?4?6???2n?22?1342(2n)2?4?6???2n1
2n?1 (方法二)因为1?1?1?2,3?3?1?4,?,2n?1?2n?1?1?4?152n2n?122n,相乘得: 2n?1
1,从而1?3?5???(2n?1)1?1?3?5???(2n?1)?. ??2?4?6???2n??2n?12?4?6???2n2n?1?? (方法三)设A=1?3?5???(2n?1),B=
2?4?6???2n2?4?6???2n,因为
3?5?7???(2n?1)2A 1. 2n?1 所以?1?3?5???(2n?1)?????2?4?6???2n?1, 2n?1从而1?3?5???(2n?1)?2?4?6???2n下面介绍几种方法证明a1?a1?a3???a1?a3?a5???a2n?1?2a?1?1 na2a2?a4a2?a4?a6???a2n(方法一)因为 2n?1?2n?1?2n?1,所以1,所以有 ?2n?1?2n?122n?1n11?31?3?5???(2n?1) ??????2k?1?2n?1?122?42?4?6???2nk?1(方法二)n?2?n?2,因为 n?2?n12n?11?n?22n?2?n,所以 1?n?2?nn?2 令n?2n?1,可以得到 ?2n?1?2n?1,所以有 n 11?31?3?5???(2n?1)??????2k?1?2n?1?122?42?4?6???2nk?1(方法三)设a?1?3?5???(2n?1),a?2n?1a所以2(n?1)an?1?an?1?(2n?1)an?an?1, nn?1n2?4?6???2n2n?2从而an?1?[2(n?1)?1]an?1?(2n?1)an,从而an?(2n?1)an?(2n?1)an?1 a1?a2?a3???an?(2n?1)an?(2n?1)an?1?(2n?1)an?1?(2n?3)an?2???5a2?3a1?(2n?1)an?3又 an?21,2n?1所以a?a?a???a?2n?1?3?2n?1?1 123n2(方法四)运用数学归纳法证明: (i)当n?1时,左边= 13?k?1n12k?1?2n?1?1 ,右边= 3?1?23?1?13?12显然不等式成立; (ii)假设n?k(k?1)时,所以要证明k?1i?1?i?1k12i?1?2k?1?1,则n?k?1时, 13?15???12k?1?12k?3?2k?1?1?12k?3, ?12i?1?2k?3?1,只要证明 2k?1?12k?3?2k?3?12k?3?2k?3?2k?1?12k?3?2k?12,这是成立的. 这就是说当n?k?1时,不等式也成立,所以,综上有a1a2? a1?a3a?a?a???a2n?1???135?2an?1?1a2?a4a2?a4?a6???a2n探究2.(2008年全国二卷)设函数f(x)? 解析:因为f(x)? 设g(x)?f(x)?ax,则 sinx.如果对任何x≥0，都有f(x)≤ax，求 2?cosxa的取值范围． sinx,所以cosx(2?cosx)?sin2x1?2cosxf'(x)??2?cosx(cosx?2)2(cosx?2)2g'(x)?f'(x)?a? , g(0)?01?2cosxcosx?2?cosx?2?1?223?a??a???a222(cosx?2)(cosx?2)cosx?2(cosx?2) 因为|cosx|?1,所以 (i)当a?1时, 3g'(x)?0231? ?????1,?2cosx?2(cosx?2)3??恒成立,即g(x)?g(0)?0,所以当a?1时, f(x)≤ax恒成立. 3 (ii)当a?0时,f(?)?1?0?a?(?),因此当a?0时,不符合题意. 222 (iii)当0?a?1时,令h(x)?sinx?3ax，则h?(x)?cosx?3a故当x??0，arccos3a?时,h?(x)?0. 3 因此h(x)在?0，arccos3a)时,h(x)?h(0)?0, arccos3a?上单调增加.故当x?(0， 即sinx?3ax.于是，当x?(0，arccos3a)时，f(x)? 所以综上有a的取值范围是?1? ?3,?????sinxsinx ??ax2?cosx3 变式：若0?xi?arccos3a,其中i?1,2,3,?,n 1且0?a?,x1?x2?x3???xn?arccos3a,求证: 3xxxx3atan1?tan2?tan3???tann?arccos3a. 22222xsinxisinxi证明：容易得到tani? ?2cosxi?12由上面那个题目知道sinxi?3axi 就可以知道tan xxx1x3a?tan2?tan3???tann?arccos3a 22222★同型衍变:(2006年全国一卷)已知函数 f(x)?1?xe?ax.若对任意 x∈(0,1) 恒有 f (x) >1, 求 a的取值范围. 1?x 解析:函数f (x)的定义域为(-∞, 1)∪(1, +∞), 导数为f?(x)?ax2?2?ae?ax. (1?x)2 (ⅰ) 当0< a≤2时, f (x) 在区间 (-∞, 1) 为增函数, 故对于任意x∈(0, 1) 恒有 f (x) > f (0) =1, 因 而这时a满足要求. (ⅱ) 当a>2时, f (x) 在区间 (-x0?12a?2, aa?2,aa?2a)为减函数, 故在区间(0, a?2a) 内任取一点, 比如取 就有 x0∈(0, 1) 且 f (x0) < f (0) =1, 因而这时a不满足要求. (ⅲ) 当a≤0时, 对于任意x∈(0, 1) 恒有 f(x)?1?xe?ax≥1?x?1, 这时a满足要求. 1?x1?x综上可知, 所求 a的取值范围为 a≤2.
高考数学 - 压轴题 - 放缩法技巧全总结（最强大）
