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数列与不等式证明的放缩
专题 数列与不等式证明的放缩
全国高考卷中经常出现与数列有关的不等式证明题,尽管不等式的证明有多种
方法
快递客服问题件处理详细方法山木方法pdf计算方法pdf华与华方法下载八字理论方法下载
,但放缩法是证明不等式的主要方法,高考出现的频率比较高,是高考的热点,2007年全国高考数学的19套理科试卷中,在不等式的证明中用到放缩法的有7套试卷,下面介绍几种与数列有关的不等式证明的放缩技巧。
一、直接使用已知的不等式定理进行放缩
a,6a,162例1、已知数列为等比数列,, a,,25n
?求数列的通项公式; a,,n
SS,nn,2S?设是数列的前项和,证明,1 (2004全国卷) na,,nn2Sn,1
aq,6,1解:?设等比数列的公比为q,则有 a,,,n4aq,1621,
a,2,1n,1 解得 ?的通项公式为 aa,23,,,nnq,3,
n,213,,na,2?由?知,q,3 ? S,,,311n,13
nnnnn,,,222231313331,,,,,,,,,,,SS,nn,2,, ?22221nn,,n,1S3231,,,n,131,,,
222nnn,,32331,,,,SS,nn,2,,1,1 即 221nn,,23231,,,Sn,1
abab,,2评注:第?小题的证明用到均值不等式定理进行放缩,要注意均值
不等式成立的条件:“正数、定值、等号”,常用的不等式定理有:?
abab,,2 ? ababab,,,,,
二、利用函数的单调性进行放缩
2例2、设函数,其中,证明对任意的正整数n,不等fxxbx,,,ln1b,0,,,,
1
111,,ln1,,,式都成立(2007山东) ,,23nnn,,
2证明:当时,函数 fxxx,,,ln1b,,1,,,,
332 令函数 hxxfxxxx,,,,,,ln1,,,,,,
2331xx,,,,12, 则 hxxx,,,,32,,xx,,11
, ?当时,,所以函数在上单调递增 x,,,0,hx,0hx0,,,,,,,,,,,
又 ?时,恒有 h00,x,,,0,hxh,,00,,,,,,,,
32 即在上恒成立 xxx,,,ln1x,,,0,,,,,
23 故当时,有 x,,,0,ln1xxx,,,,,,,
1 对任意正整数,取 nx,,,,0,,,n
111,,ln1,,, 则有 所以原不等式成立。 ,,23nnn,,
评注:本题运用函数的单调性进行放缩,本题有较高的难度,难就难在要从放
缩的不等式中去发现要构成的函数,清楚了这一点,后面的难度就不高
了。
三、增加或舍去某些项进行放缩
aa例3、已知数列中的相邻两项,是关于的方程 xa,,21k,2kn
2kkxkxk,,,,32320,的两个根,且 aak,,1,2,3,,,,,212kk,
aaaa (?)求,,,; 1357
S (?)求数列的前项和; a2n,,2nn
,,sinn1 (?)记 fn,,3,,,,2sinn,,
2
ffffn2341,,,,,,,,,,,,,1111,,,,,,,, T,,,,,naaaaaaaa,123456212nn,15, (2007浙江) 求证:,,,TnN,,n624
2kkkxk,3xkxk,,,,32320,解:(?)方程的两个根为,, x,2,,12
x,3x,2a,2 当时,,,所以; k,1121
x,6x,4a,4 当时,,,所以; k,2123
x,9x,8a,8 当时,,,所以; k,3125
x,12x,16a,12 当时,,,所以。 k,4127
Saaa,,,, (?) 2122nn
2n,,,,,,,,363222n ,,,,
233nn,n,1,,,22 2
fn,1,,,1,,111T (?)证明:,,,,, naaaaaaaa,123456212nn,
11115T,,T,,, 所以 12aa6aaaa24121234
fn,1,,,1,,111T,,,,,时, 当n,3naaaaaa63456212nn,
,,1111 ,,,,,,,aaaaaa63456212nn,,,
11111111,,,,,,,,,, ,,23nn,662622,6626,,
fn,1,,,1,,511T,,,,, 同时, naaaaaa245678212nn,
3
,,5111 ,,,,,,,aaaaaa245678212nn,,,
51111515,,,,,,, ,,,,,34nn,2492922,249224,,
15,时, 综上,当nN,,,Tn624
15评注:第(?)小题的证明必须注意到,,从上面的证明过程中,T,T12624
我们可以发现,必须勇敢地有依据地放弃某些负数项整个值就会增大,
放弃某些正数项整个值就会缩小,这种方法是很独特的。
1a,2aan,,,例4、设数列满足, 1,2,3,a,,,,1nn,1nan
?证明:对一切正整数成立;(2004重庆卷) nan,,21n
an1,2,3,bbbn,,?令,判断与的大小,并说明理由。 ,,nn,1nn
?证法一、当时,,不等式成立 a,,,,2211n,11
假设时,成立 ak,,21nk,k
1122aakk,,,,,,,,,, 那么当时,2212211 nk,,1,,kk,122aakk
ak,,,211 ?时,成立 nk,,1,,k,1
综上,由数学归纳法可知:对一切正整数n成立。 an,,21n证法二、当时,,结论成立 a,,,,,23211n,11
假设时结论成立,即 ak,,21nk,k
1 那么当时,由函数单调递增和归纳 fxxx,,,1nk,,1,,,,x
11aak,,,,,21 假设有 kk,1ak,21k
4
1 因此只需证2123kk,,,,
21k,
21,, 就是证: 2123kk,,,,,,21k,,,
1 即证21223kk,,,,,21k,
上式显然成立 所以时,结论成立。 nk,,1
综上,由数学归纳法可知,对一切正整数均成立。 nan,,21n
证法三、由递推公式得
112222aa,,,aa,,, 22nn,nn,,11222aan,n,12
122 „„ aa,,,2 212a1
上述各式相加并化简得
111222aann,,,,,,,,,, 21221 ,,,,n1222aaa,n121
,,,,,22212nnn,,
又时,显然成立 an,,21n,1n
所以对一切正整数n均成立 an,,21n
bb,?解: nn,1
,,ban11nn,,nn,,11 证明一: 11,,,,,,,,,,2ban21,annn,,,111,,nn,,n
22111,,,,n,,n,,,,,21nn,21nn,,,,,242,,,,, ,,,,11121n,211nn,,,,n,n,22
bb,b,0又,故 nnn,1
5
,,aaa11nn,1n,,, 证明二:bb ,,,a,,nn,1na1,nnnn,1n,,
112,,,, ,,,,nnna,,,,,nnnn1121,,,,,,n,,,,nna,nna,11,,,,nn
,,nnnn,,,,121,,,,nnn,,,11,,,,,, ,,
nnnna,,,11nnnna,,,11,,,,,,,,nnnn,,1bb, 所以 ,,,00a,,nn,1nnnna,,1,,n
22aa22nn,1,,,证明三:bb nn,11,nn
22,,,,aa11112nn ,,,,a2,,,2,,,,n22nan,1nan,1nn,,,,
1121n,111,,,,,,,2,,,0 ,,,,nnn,,121nnn,,121,,,,
22bb,b,0 ? 又 所以 bb,nnn,1,nn1
评注:第?小题证法一、证法三、第?小题证明一均用到舍去某些项进行放缩,
第?小题证法二用到函数单调性进行放缩。
四、对分式中的分子、分母进行放缩
S21Sa,,例5、已知正数数列的前n项和为,且对任意正整数n满足 a,,nnnn
?求数列的通项公式; a,,n
11Tb,?设,数列的前n项和为,求证:; bT,,,nnnnaa2nn,1
11171,,,,,?求证: 222aaan,6221,,12n
241Sa,,21Sa,,解:?把两边平方得 ,,nnnn
6
2? ,41Sa,,,,nn, ? ,241Sa,,,,,,,nn11, ?
22 ?—?得 ,aaaa,,,,20422aaaaa,,,,,,,,nnnn,,11nnnnn,,,111
,aa,,,20aanN,,,2 ?数列是正数数列 ?即 a,,,,nn,1nn1,n
a,1由已知得即 , 21Sa,,210aaa,,,,,111111
an,,21?数列是首项为1,公差为2的等差数列 ? a,,nn
11111,,? b,,,,n,,aannnn,,,,212122121,,,,,,,1nn
,,111111,,,,,, ? T,,,,,,,1n,,,,,,,,nn,,23352121,,,,,,,,
111111,,,,1,,, 即 T,,,n221n,22212n,2,,,,
an,,21?由?知 n
1111111,,, ? ,,,,,2222222aaa13521n,,,n12
111,,,,,1 35572121,,,,nn,,,,
1111111,,,,,,,, ,,,,,,,,1,,,,,,,,235572121nn,,,,,,,,,,
1111171,,,,,1,,,1,, ,,6221n,6221n,2321n,,,,,,,
评注:第?小题用到裂项放缩和舍去某些项进行放缩,第?小题用到对分母进
行放缩,通过放缩把陌生的数列求和问题转化为熟悉的数列求和问题。
11111111,,,,,,常见的放缩有,, 22nnnnn,,11nnnnn,,11,,,,
7
222,等等。 nnnn,,,,,111nnnnn,,,,,,2211,,,,
五、利用二项式定理进行放缩
n,1San,,,,21221、已知数列的前项和为,且 na,,,,nnn
1111求证:当时,有,,,, n,3SSS1034n
21n,Snn,,,,,,,,,,352722122证明: ,,n
211nn,,Tnn,,,,,,,,,,,,35272212212 令 ? ,,,,n
21nn,23252212212Tnn,,,,,,,,,,, 则 ? ,,,,n
21nn,,,,,,,,,,Tn32222212 ?,?得 ,,,,n
n,1212,,,nn,,,,1221n n,,,,,32212,,,,12,
nnSTnn,,,,,,22121Tn,,,,2121 ? ? ,,,,,,nnn
nnnn011,211,,,,,,,CCCC? ,,nnnn
nnn011,?当时, 221,,,,,,CCCCnn,3,,nnnn
n? ? 2121,,,nSnn,,,2121,,,,n
11111,,,,,? ,,Snnnn,,,,212122121,,,,,,n
1111111111,,,,,,,,,,,? ,,,,,,,,,,,,,,,SSS257792121nn,,,,,,,,34n,,111111,, ,,,,,,,25212510n,,,
1111,,,,?时,有成立 n,3SSS1034n
8
an2,,aa,1nnb,,12、已知正项数列的前项和S, nanN,,,,,,,nnn,a22n,,
3 (1)求数列的通项公式 (2)求证: a,,b2,,nn2
2aa,11aS,,a,1a,0(1)解:时,或 ,n,111112
a,1 ?由于是正项数列 ? a,,1n
22aaaa,,nnnn,,11 当时,aSS,,,, n,2nnn,122
整理得 aaaaaa,,,,,,,,nnnnnn,,,111
aa,,1a,0 ? ? nnn,1
?数列是首项为1,公差为1的等差数列 a,,n
ann,,,,,111? ,,n
a,1又时,适合上式 ? annN,,n,1,,1n,
n1,,b,,(2)证明:由(1)知 1n,,n2,,
n12111,,,,,,012,,,,,,CCC? 1,,,,,,nnnnnn222,,,,,,
rn11,,,,rn,,,CC ,,,,nnnn22,,,,
rnnnr,,,11,,,,11,,rC,,而 n,,rr2!2nrn,,,
nnnr,,,1111111,,,,,,,,, rrrr!22nnnr!2
9
nn21111,,,,,,? b,,,,,,,11,,,,,,nn2222,,,,,,
1,,11,,n,,n,112,,,, ,,,,22,,12,,,12
11133,,01 又? ? bCC,,,,,1,,b2nnn,,nn2222,,
评注:利用二项式定理进行放缩,主要方法是舍去某些项或对分式中的分子、
分母进行放缩,放缩的目的是使所得的数列转化为熟悉的、较容易的数
列求和类型,从而达到证明命题的目的。
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11,,aan,,,,11a,11、数列满足且 a,,,,nn,11,,n2nnn,2,,
?用数学归纳法证明: an,,22,,n
2?已知不等式对于恒成立,证明: ln1,,xxae,x,0,,n
11,,aa,,,,,122证明:?成立; 21,,22112,,,
?假设时,有成立,那么时, ak,,22nk,nk,,1,,k
11,,aaa,,,,,12 kkk,1,,2kkk,2,,
?结论也成立 nk,,1
根据??得,恒有 an,,22,,n
1111,,,,aaaa,,,,,,,,,,121?? nnnn,1,,,,2121nn,,nnnn,,22,,,,
10
11,,a,,,1, , e,1n,,21n,nn,2,,
对于恒成立 ln1,,xxx,0,,
1111,,aaalnlnln1ln,,,,,,,? nnn,1,,2121nn,,nnnn,,22,,
111? aa,,,,lnlnnn,1n,1nn,12
?时, lnlnlnlnlnaaaaa,,,,,,,,lnlnlnaaan,2,,,,,,nnnnn,,,112211
111111111,,,,,,,,,,,,,,,, ,,,,,,23nnn,12223212,,,,,,
11,,1,,,n,11111342,,2 ?成立。 ae,,,,,,,,,,112nn1nn222,12
从这个例子可以看出逐差、逐商法本质上是用来对数列的项进行放大或缩
小处理用的。
2a,32、设数列满足,1、2„„,已知 n,aaana,,,1,,1nnnn,1
1111an,,2,,,,?证明; ? n,,,aaa111212n证明?用数列归纳法证明:
a,,,312i)时,不等式成立; n,11
ak,,2ii)假设有成立。 nk,n
k,,2,,, 那么,时,在递增 fxxkx,,,1nk,,1,,,,2,,
2akkkkk,,,,,,,,,,2212512 ? ,,,,k,1
an,,2an,,2?时,不等式也成立 综上,恒有成立 nk,,1nn
11
11,,1,,,n11111142,,,,??成立 ,,,,,,1,,1nn,1482222,,1,2
n,114,,a?只要证明就可以,其中 122,,,an,,1n
2aan,,2,,1222,,,,aanaann,nnnn1? ,,,,21111,,,,aaaannnn
,,11aa,1ann,,11nn,12,,,,,,,,, 时,, 11212aaan,1,,,,n11,,,111aaann,,121
n,111,,n,1 从而有成立 ? ,12,,a,,na12,,,n
n,11111111,,?成立 ,,,,,,,,,,aaa1114822,,,,,12n
?原不等式成立。
从这个例子可以看出,条件与目标有的时候相差很远,需要我们联想才能
121n,,,,,,,qqq1联系起来。如当时就有,特别地 q,1,q1
n,11111,,,,等。 121,,,,,,,,,,n2422,,,,
112,3、已知数列, aaanN,,,a,,,0nnn,,1122n
111n,1,, 求证:? ? ,,ann2aann,2nn,1
12aa,,0证明:?由知 aaa,,nn,1nnn,,112n
112 ? aaaaaa,,,,nnnnnn,,,,111122nn
111,,, 2aannn,1
12
,,,,,,11111111? ,,,,,,,,,,,,,,aaaaaaaa001121nnn,,,,,,,
111 ,,,,22212n
111,,,,,1 12231,,,,nn,,
111111,,,,,,,,,,,,,,11 ,,2,,,,,,2231nn,n,,,,,,
111a,0,,, 又, ? 22a,,,ann0nan2n
2nnn,,,1112an,,1 又则,,,aaa aaa,,n,1nnn,,,111nnn,,11222nnn
2n,,aa nn,121,,nn
21n,,,aaaa ? nnnn,,1122,,nnn1
111111,,,,,? 22aannnnnn,,,,11nn,1
,,,,,,11111111? ,,,,,,,,,,,,,,aaaaaaaa001121nnn,,,,,,,
1111111,,,,,,,,,,,,, ,,1,,,,,,12231nn,n,1,,,,,,
11112n,a,221,,,,,,,1又 ? 0an,1ann,,11nn
n,1a,0 又 ? a,nnn,2
n,1综上,成立。 ,,annn,2
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3na3,n,14、已知数列满足,且,,, (2006江西) annN2,aa,,,,,nn1,,21an2n,1
?求数列的通项公式; a,,n
aaan,,,,,2!?证明:对一切正整数,不等式恒成立 n12n
,,3nann11,n,1,解:?由得 a11,,,,,n,,21anaa3n,1nn,1,,
,,11n11,, ?是首项,公比为的等比数列 1,,,a3a31n,,
n,nn13, ? ,,,,,1anN,,nnn,a331n
nnn,3?证明:由?知 a,,nn131,1,n3
n! ?aaa,,,, 12n111,,,,,,111,,,,,,,,,2n333,,,,,,
1111,,,,,,111,,,,aaan,,,,,2! ?要证 只需证: 12n,,,,,,2n3332,,,,,,
下面用数字归纳法证明:
111111,,,,,,,, 1111,,,,,,,,,,,,,,,,,,,22nn333333,,,,,,,,
1212?当时,左边,右边 ,,,1,,,1n,13333
不等式显然成立 ,,,
?假设时,不等式成立 ,nk,,,
111111,,,,,,,,1111,,,,,,,,即 ,,,,,,,,22kk333333,,,,,,,,
14
1111,,,,,,,,1111,,,,则当时 nk,,1,,,,,,,,21kk,3333,,,,,,,,
,,1111,,,, ,,,,,,11,,,,21kk,,,3333,,,,,,
11111111,,,,,,,,,,,,,,1 ,,,,2112kkkk,,33333333,,,,
1111111,,,,,,,,,,,,,1 ,,,,2112kkk,,3333333,,,,
111,,,,,,,1 ?当时,不等式也成立 ,nk,,1,,,,21k,333,,
, 根据??可知,对一切 nN,
111111,,,,,,,,1111,,,,,,,, 不等式都成立 ,,,,,,,,22nn333333,,,,,,,,
111111,,,,,,,,1111,,,,,,,, ? ,,,,,,,,22nn333333,,,,,,,,
11,,1,,,n11111133,,,, ,,,,,,,,,111,,nn1223223,,1,3
aaan,,,,,2!?对一切正整数,不等式恒成立 n12n
b,4bba,6baa5 .在数列、中,已知,,且、、成等比数列,、、ab,,,,11nn,1n,1nnnn
,a成等差数列,() nN,n,1
aaabbb?.求、、及、、,由此猜想、的通项公式,并证明你的结论; ab,,,,234234nn11117,,,,,,,,?.证明:. abababab,,,,20112233nn
15
,bbbaaa解:?.由已知、、成等比数列,、、成等差数列,() nN,nn,1n,1n,1nn
22baa,,b,4a,6, ,,,代入计算得: ?Qabb,,nnn,,1111,nnn1
a,12a,20a,30,,, 234
b,16b,25b,9,,, _______3分 234
,2ann,,,(1)(2)()nN,由此猜想, ,_______5分 bn,,(1)nn
证明:(1)当,由上面计算知猜想的结论成立; n,1
,2akk,,,(1)(2)kkN,,1,(2)假设当时结论成立,即,, bk,,(1)nk,,,kk
222a(1)(2)kk,,22k则当时,由于, ?abb,,nk,,1bk,,,,,(1)1,,,,k1kkk12bk(1),k
2?当时,结论成立 _______7分 bn,,(1)nk,,1n
2又 abakkkkk,,,,,,,,,,22(2)(1)(2)(2)(3)kkk,,11
ann,,,(1)(2)当时,也成立 ?,,,,,(1)1(1)2kknk,,1,,,,n
,2ann,,,(1)(2)由(1)(2)所证可知对任意的自然数,结论,都nN,bn,,(1)nn
成立 _______9分
1117,,,?.因为 _______10分 ab,,64102011
2当时,由 abnnnnnnn,,,,,,,,,,,(1)(2)(1)(1)(23)2(1)n,2nn
11111,,,() _______11分 abnnnn,,,2(1)21nn
11111111111,,,,,,,,,,,,,,,,,()abababnn,,,,102233411122nn
1111117 _____13分 ,,,,,,()1022110420n,
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