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文科数学真题模拟汇编14:数列
广东高考文科数学真题模拟汇编
14:数列
*,,SS,2a,1a1((2009广州一模文数)已知数列的前项和为,对任意N都有, nn,nnnn
,,aaa,则的值为 ,数列的通项公式 . nn1
n,1121(;
n,22. (2010广州二模文数)图2是一个有层的六边形点阵.它的中心是一个点,算作第一层, n,,
第2层每边有2个点,第3层每边有3个点 ,„,第层每边有个点, nn
则这个点阵的点数共有 个. 图2
2331nn,,2.
3((2010广州一模文数)如图3所示的三角形数阵叫“莱布尼兹调和三角形”, 1 它们是由整数的倒数组成的,第行有个数且两端 nn1
111n?2的数均为,每个数是它下一行左右相邻两数 ,, n22
111111111111的和,如,,,„, ,,,,,, 1222363412363
则第7行第4个数(从左往右数)为 1111 11441212A( B( 1401051111 112020530C( D( 图2 42601 3、答案A 5
„„„„„„„„„„„„„„a,1q,2a,64a4((2010广州一模文数)在等比数列中,,公比,若,则的值 n,,1nn„
为 (
4(答案7
2a,3aa5((2011广州一模文数)已知等比数列的公比是,,则的值是 . ,,35n
125(答案
naaaa,,,,,an,,,11a6((2011广州二模文数)已知数列的通项公式是,则 ,,,,,,12310nn
,55,5A( B( C(5 D(55 6、答案C
7((2012广州一模文数)两千多年前,古希腊毕达哥拉斯学派的数学家曾经在沙滩上研究数学问题,他们在
沙滩上画点或用小石子来
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
示数,按照点或小石子能排列的形状对数进行分类,如图2中的实心点个数1,
a,1a,55,12,22,„,被称为五角形数,其中第1个五角形数记作,第2个五角形数记作,第3个五12
a,12a,22a,a,145角形数记作,第4个五角形数记作,„,若按此规律继续下去,则 ,若,345n则 ( n,
22 5 12 1
图2 7(答案35,10
(n,3)13、 (2005广东)设平面内有条直线,其中有且仅有两条直线互相平行,任意三条直线不过同一n
f(n)f(4)f(n),n,4点(若用表示这条直线交点的个数,则=____________;当时, ((用n
表示) n
113【答案】5, (n,1)(n,2)2
f(4),5f(3),2f(4),5f(5),9解:由图B可得,由,,,
图B f(6),14(n,1)f(n),2,3,4,?,(n,1),可推得?n每增加1,则交点增加个,?
(2,n,1)(n,2)1,( ,(n,1)(n,2)22
14、(2006广东)已知等差数列共有10项,其中奇数项之和15,偶数项之和为30,则其公差是 A.5 B.4 C. 3 D.2
5a,20d,15,114、,故选C. ,d,3,5a,25d,301,
15、(2006广东)在德国不莱梅举行的第48届世乒赛期间,某商场橱窗里用同样的乒乓球堆成若干准“正三棱锥”形的展品,其中第一堆只有一层,就一个乒乓球;第2、3、4、„堆最底层(第一层)分别按图4所示方式固定摆放.从第一层开始,每层的小球自然垒放在下一层之上,第n堆第n层就放一个乒乓球,f(n)f(3),f(n),以表示第n堆的乒乓球总数,则 ; (答案用n表示) .
n(n,1)(n,2)f(3),15、10,f(n), 6
2a,Snn,,916((2007广东文数)已知数列a的前项和,则其通项 ;若它的n,,nnn
k58,,ak,第项满足,则 ( k
16. 2n-10
17((2007广东文数)在一个袋子中装有分别标注数字1,2,3,4,5的五个小球,这些小球除标注的数
字外完全相同(现从中随机取出2个小球,则取出的小球标注的数字之和为3或6的概率是( )
3111,( ,( ,( ,( 1010125
17.A
SS,,4,20Sd,18. (2008广东文数)记等差数列的前项和为,若,则该数列的公差( ) n24n
A、2 B、3 C、6 D、7
SSSdd,,,,,,412318【解析】,选B. 422
2aa{a}aaa19 (2009广东文科)已知等比数列的公比为正数,且?=2,=1,则= 1n3925
212A. B. C. D.2 22
2842qaqaqaq,,2q,2{a}19【答案】B【解析】设公比为,由已知得,即,因为等比数列的公比为正,,n111
a122q,2数,所以,故,选B a,,,1q22
20、(2011•广东文数)已知{a}是递增等比数列,a=2,a,a=4,则此数列的公比q= 2 ( n2432220\解答:解:?{a}是递增等比数列,且a=2,则公比q,1又?a,a=a(q,q)=2(q,q)=4 n24322即q,q,2=0解得q=2,或q=,1(舍去)故此数列的公比q=2故答案为:2
12{a}aaa,211((2012广东文数)若等比数列满足,则_______________( aa,n135242
121: 4
22. (2009广州一模文数) (本小题满分14分)
n2*,,a,a(n,a)a,1x,2x,b,0已知数列的相邻两项是关于的方程N的两根,且. xnnn,11n
1,,n,,(1) 求证: 数列是等比数列; a2,,n3,,
*S,,,b,S,0a,(2) 设是数列的前项和, 问是否存在常数,使得对任意N都成立,若存在, nn,nnnn
,求出的取值范围; 若不存在, 请说明理由.
22((本小题满分14分)
(本小题主要考查数列的通项公式、数列前项和、不等式等基础知识,考查化归与转化、分类与整合、n
特殊与一般的数学思想
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
,以及推理论证能力、运算求解能力和抽象概括能力)
n2*a,a(n,)x,2x,b,0 (1)证法1: ?是关于的方程N的两根, xnn,1n
n,a,a,2,nn,1 ? ,b,aa.nnn,1,
11,,nn,1na,a,2 由,得, a2a2,,,,,,,,,1nnn,1n33,,
211,,n,1,,故数列是首项为,公比为的等比数列. a2a,,,,1n333,,
n2*a,a(n,)x,2x,b,0 证法2: ?是关于的方程N的两根, xnn,1n
n,a,a,2,nn,1 ? ,b,aa.nnn,1,
1,,11nnnn,1,1,,,2a,,a,,22,a,,2nnn,13,,33,,,1?, ,111nnna,,2a,,2a,,2nnn333
211,,n,1,, 故数列是首项为,公比为的等比数列. a2a,,,,1n333,,
111n,1nnn(2)解: 由(1)得, 即,,. ,,,,a,,2,,,1a,2,,1nn333
1nn,1nn,1?,,,, ,,,,b,aa,2,,1,2,,1nnn,19
1n2n,1 ,,. ,,,2,,2,19
S,a,a,a,?,a? n123n
12n23n ,,,, ,,,,,,,,,2,2,2,?,2,,1,,1,?,,13
n,,,,1,1,1n,1,2,2, . ,,32,,
*b,S,0,要使对任意N都成立, n,nn
n,,,,,,1,11n*n,12n,1,2,2,,0即(*)对任意N都成立. ,,,,2,,2,1n,,,329,,
1,2n,1nn,1? 当为正奇数时, 由(*)式得, ,,2,2,1,,,2,1,0n931,n,1nn,1即, ,,,,2,12,1,,,2,1,039
n,12,1,0?,
1n?对任意正奇数都成立. ,,,,2,1n3
1n1n,1当且仅当时, 有最小值. ,,2,13
,,1?.
1,2n,1nn,1? 当为正偶数时, 由(*)式得,,, 2,2,1,,,2,2,0n9312,n,1nn即, ,,,,2,12,1,,,2,1,039
n2,1,0?,
1n,1?对任意正偶数都成立. ,,,,2,1n6
13n,1n,2当且仅当时, 有最小值. ,,2,126
3,,?. 2
*,b,S,0,,,,,,,1综上所述, 存在常数,使得对任意N都成立, 的取值范围是. n,nn
23((2010广州一模文数)(本小题满分14分)
2*333a,0n,Naaaaaa,,,,,,,a已知数列满足对任意的,都有,且( ,,,,nn1212nn
aa(1)求,的值; 12
aa(2)求数列的通项公式; ,,nn
,,11S(3)设数列的前项和为,不等式对任意的正整数恒成立,求实 Sa,,log1nn,,,,nnaaa3nn,2,,
数的取值范围( a
23((本小题满分14分)
(本小题主要考查数列通项、求和与不等式等知识,考查化归与转化的数学思想方法,以及抽象概括能力、
运算求解能力和创新意识)
32n,1aa,(1)解:当时,有, 11
a,1a,0由于,所以( 1n
233n,2aaaa,,,当时,有, ,,1212
a,1a,2a,0将代入上式,由于,所以( 12n
2333aaaaaa,,,,,,,(2)解:由于, ? ,,1212nn
23333aaaaaaaa,,,,,,,,,则有( ? ,,,,121121nnnn
223aaaaaaaa,,,,,,,,,?,?,得, ,,,,,,nnnn112112
2a,0aaaaa,,,,,2由于,所以( ? ,,nnnn,,1121
2aaaaa,,,,,2n?2同样有, ? ,,,,nnn,121
22aaaa,,,?,?,得( ,,nnnn11
aa,,1所以( nn,1
aa,,1aa,,1n?1a由于,即当时都有,所以数列是首项为1,公差为1的等差数列( ,,nn,121n
an,故( n
11111,,an,(3)解:由(2)知,则( ,,,n,,aannnn,,222,,,,nn,2
11111所以 S,,,,,,naaaaaaaaaa132435112,,,nnnn
11111111111111,,,,,,,,,, ,,,,,,,,,,,1,,,,,,,,,,2322423521122nnnn,,,,,,,,,,,,,
11113111,,,, ( ,,,,1,,,,,,,2212nn,,4212nn,,,,,,
1S,?数列单调递增( ?SS,,,0,,nnn,1nn,,13,,,,
1所以( SS,,,,n1min3
111要使不等式对任意正整数恒成立,只要( Sa,,,,alog1nlog1,,,,naa333
10,,a01,,a?,?(
11,,aa?,即( 0,,a2
1,,所以,实数的取值范围是( 0,a,,2,,
24((2011广州一模文数)(本小题满分14分)
*San,,,1(){a}S{b} 已知数列的前项和为,且满足N.各项为正数的数列中,对于一切nnnnnn
n1n*bbb,,,1,2,3N,有, 且. ,n,123,,,bbbb,1k,,111kkn
{}a{}b (1)求数列和的通项公式; nn
{}abTT,2 (2)设数列的前项和为,求证:. nnnnn24((本小题满分14分)
(本小题主要考查数列、不等式等知识, 考查化归与转化、分类与整合的数学思想方法,以及抽象概括能
力、推理论证能力、运算求解能力和创新意识)
Sa,,1(1)解:?, nn
1aSa,,,1n,1当时,, 解得. „„1分 a,11112
n,2aSS,,,,,,11aa当时,, ,,,,nnn,1nn,1
a1n2aa,得, 即. „„ 3分 ,nn,1a2,1n
11{a}?数列是首项为, 公比为的等比数列. n22
n,1111,,?. „„ 4分 a,,,,,nn222,,
n1n* ? 对于一切N,有, ? ,n,,,,bbbb,1k,,111kkn
n,111n,n,2当时, 有 , ? ,,bbbb,,k,1kkn,11
11nn,? ? 得: ,,,
bbbbbb,,,nnnn,,1111
(1)0nbnbb,,,,化简得: , ? nn,11
nbnbb,,,,(1)0n,1用替换?式中的,得:, ? „„6分 nnn,,211
bbbb,,,?,? 整理得:, nnnn,,,211
n,2{}b ?当时, 数列为等差数列. n
bbbb,,,,1?, 3221
{}b? 数列为等差数列. „„ 8分 n
bb,,1,2 ? 12
{}bd,1 ?数列的公差. n
bnn,,,,11 ?. „„ 10分 ,,n
{}abT(2)证明:?数列的前项和为, nnnn
123n ?, ? T,,,,,n23n2222
112n ? , ? ,,,,Tn221,n2222
1111n?,?得: „„ 12分 ,,,,,Tn21,nn22222
n,,11,,1,,,,,22,,n,,,, ,,,1n121,2
n,2 . ,,1n,12
n,2?. „„14分 T,,,22nn2
25((2011广州二模文数)(本小题满分14分)
S,55SS,210已知等差数列,a,的前项和为,且,( nn10n20
a(1)求数列的通项公式; ,,n
a*nkmkm,,,2,,Nbkbb(2)设,是否存在、,使得、、成等比数列(若存在,,mb,,1mknan,1
k求出所有符合条件的、的值;若不存在,请说明理由( m
25((本小题满分14分)
(本小题主要考查等差数列、等比数列、不等式等基础知识,考查方程思想以及运算求解能力()
nn,1,,da解:(1)设等差数列的公差为,则(„„„„„„„„„„„„„„„1分 Snad,,,,nn12
109,,1055,ad,,1,,2由已知,得„„„„„„„„„„„„„„„„„„„„„„„„„„„3分 ,2019,,20210.ad,,1,,2
2911,ad,,a,1,,,11即解得„„„„„„„„„„„„„„„„„„„„„„„„„„„„5分 ,,21921.ad,,d,1.,1,
,aandn,,,,(1)n,N所以()(„„„„„„„„„„„„„„„„„„„„„„„„6分 n1
bkbbkmmk,,,2,,N(2)假设存在、,使得、、成等比数列, m,,1mk2bbb,则(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„7分 mk1
ann因为,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„8分 ,,bn1,an,1n
1mk所以( bbb,,,,,1mk211mk,,
2mk1,,所以(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„9分 ,,,,mk,,121,,
22m整理,得(„„„„„„„„„„„„„„„„„„„„„„„„„„„„10分 k,2,,,mm21
k以下给出求~的三种方法: m
2k,0,,,,mm210方法1:因为,所以(„„„„„„„„„„„„„„„„„„„„„11分
1212,,,,m解得(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„12分
*mm,,2,N因为,
m,2k,8所以,此时(
bm,2k,8bb故存在、,使得、、成等比数列(„„„„„„„„„„„„„„„„„14分 1mk
22mkm,方法2:因为,所以(„„„„„„„„„„„„„„„„„„„11分 km,,2,,,mm21
2m,12m即,即( ,0,,1022mm,,21mm,,21
,,,,112m112,,,m解得或(„„„„„„„„„„„„„„„„„„„„„„„„12分
*mm,,2,N因为,
m,2k,8所以,此时(
bm,2k,8bb故存在、,使得、、成等比数列(„„„„„„„„„„„„„„„„„14分 1mk
22mkm,,2方法3:因为,所以(„„„„„„„„„„„„„„„„„11分 k,,22,,,mm21
22m221mm,,即,即( ,,10,022mm,,21mm,,21
13,13,解得或(„„„„„„„„„„„„„„„„„„„12分 12,,,m,,,m1222
*mm,,2,N因为,
m,2k,8所以,此时(
bm,2k,8bb故存在、,使得、、成等比数列(„„„„„„„„„„„„„„„„„14分 1mk
26((2012广州一模文数)(本小题满分14分)
d,0SS,70aaaa已知等差数列的公差,它的前项和为,若,且,,成等比数列( n,,n52722n
a(1)求数列的通项公式; ,,n
,,113T(2)设数列的前项和为,求证:( ?T,n,,nnS68n,,
26((本小题满分14分)
(本小题主要考查等差数列、等比数列、裂项求和等知识,考查化归与转化的数学思想方法,以及抽象概
括能力、运算求解能力和创新意识)
a(1)解:因为数列是等差数列, ,,n
nn,1,,aand,,,1所以,(„„„„„„„„„„„„„„„„„„„„1分 Snad,,,,n1n12
51070,ad,,,S,70,,1,,5依题意,有即„„„„„„„„„„„„„„„3分 ,,22aaa,.adadad,,,,621.,,,,,,,,7222,111,
a,6d,4解得,(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„5分 1
*an,,42n,Na所以数列的通项公式为()(„„„„„„„„„„„„„„„„„„„6分 ,,nn
2Snn,,24(2)证明:由(1)可得(„„„„„„„„„„„„„„„„„„„„„„„„„„7分 n
111111,,,,所以(„„„„„„„„„„„„„„„„„„„8分 ,,,,2Snnnn,,42nn,2422,,,,n
11111所以 T,,,,,,LnSSSSS1231,nn
11111111111111,,,,,,,,,, „„„„„9分 ,,,,,,,,,,,1,,,,,,,,,,4342443541142nnnn,,,,,,,,,,,,,
1111,, ,,,,1,,4212nn,,,,
3111,, („„„„„„„„„„„„„„„„„„„„„„„„„„„10分 ,,,,,8412nn,,,,
31113,,因为,所以(„„„„„„„„„„„„„„„„„„11分 T,,,,,T,0nn,,nn,,84128,,
111,,T因为,所以数列是递增数列(„„„„„„„„„„„„12分 TT,,,,0,,nnn,1,,nn,,413,,
1所以(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„13分 TT,,n16
13 所以(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„14分 ,,Tn68
*Sa,0{}anN,27. (2012广州二模文数(本小题满分)14分)已知数列的前项和为,对任意,都有nnnn
且
(1)(2)aa,,lnannn,1,令. ,S,bnn2lnan
{}a(1)求数列的通项公式; n
*bbb,,,…[1,2012]kkN(),(2)使乘积为整数的叫“龙数”,求区间内的所有“龙数”之和; 12k
bb(3)判断与的大小关系,并说明理由。 nn,1
27((本小题满分14分)
(本小题主要考查数列、不等式等知识, 考查化归与转化、分类与整合的数学思想方法,以及抽象概括能
力、推理论证能力、运算求解能力和创新意识)
2aa,,12,,,,aa,,2nnnn(1)解:由于, S,,n22
2aa,,211n,1 当时,, „„„„„ 1分 aS,,112
2aa,,,20 整理得, 11
a,,1a,2 解得或. 11
a,0 ?, n
a,2 ?. „„„„„ 2分 1
22aaaa,,,,22nnnn,,11n,2aSS,, 当时,, „„„„„ 3分 ,,nnn,122
22aaaa,,,,0 化简得, nnnn,,11
aaaa,,,,10 ?. ,,,,nnnn,,11
a,0 ?, n
aa,,1 ?. „„„„„ 4分 nn,1
21a ?数列是首项为,公差为的等差数列. ,,n
ann,,,,,211 ?. „„„„„ 5分 ,,n
ln2n,,,lnan,1(2)解:?, ,,bnln1n,,,lnan
ln2k,,,ln3ln4bbb,,,,,, ? ,12kln2ln3ln1k,,,
ln2k,,, ,ln2
log2k, . „„„„„ 6分 ,,,2
mkm,,22(log2k, 令,则为整数), „„„„„ 7分 ,m,,2
mm1222012,,,322014,, 由,得, m,2,3,4,,10 ?.
2310k22,22,,22,,,1,2012 ?在区间内的值为, „„„„„ 8分 ,,
2310222222,,,,,, 其和为 ,,,,,,
2310,,,,,,22229 ,,
29212,,,, „„„„„ 9分 ,,1812,
,2026 . „„„„„ 10分
ln2ln1nn,,,,,,3)解法1:?, (b,,,1nln1ln1nn,,,,,,
ln3n,,,
ln3ln1nn,,ln2n,,,,,,,bn,1, ? „„„„„ 11分 ,2ln2n,,,ln2n,b,,n
ln1n,,,
2ln3ln1nn,,,,,,,,,
,,2,, „„„„„ 12分 ,2ln2n,,,
2,,ln31nn,,,,,,,,, 24ln2n,,,
22,,nn,,,31,,ln,,,,2,,,,,, „„„„„ 13分 ,24ln2n,,,
,1 .
bb, ?. „„„„„ 14分 nn,1
ln2ln1nn,,,,,,解法2:?, b,,,1nln1ln1nn,,,,,,
ln3ln2nn,,,,,, ? bb,,,nn,1ln2ln1nn,,,,,,
2ln3ln1ln2nnn,,,,,,,,,, „„„„„ 11分 ,ln2ln1nn,,,,,,
2,,ln3ln1nn,,,,,,,2,,ln2n,,,,2,, „„„„„ 12分 ,ln2ln1nn,,,,,,
2,,ln31nn,,,,,,2,,ln2n,,,,2,, ,ln2ln1nn,,,,,,
22,,131nn,,,,,2lnln2,,n,,,,,,22,,,,,, „„„„„ 13分 ,,,,,ln2ln1nn,,
,0 .
bb, ?. „„„„„ 14分 nn,1
ln1x,,,解法3:设, fxx,,2,,,,lnx
11lnln1xx,,,,'xx,1 则. „„„„„ 11分 fx,,,2lnxx,2 ?,
1111 ?. lnln1lnln10xxxx,,,,,,,,,,xxxx,1
'fx,0 ?. „„„„„ 12分 ,,
fx2,,, ?函数在上单调递减. ,,,,
* ?N, n,
212,,,,nn ?.
fnfn,,,21 ?. ,,,,
ln3ln2nn,,,,,, ?. „„„„„ 13分 ,ln2ln1nn,,,,,,
bb, ?. „„„„„ 14分 nn,1
2q(0,q,1){a}{a}28、(2006广东)已知公比为的无穷等比数列各项的和为9,无穷等比数列各项nn
81的和为. 5
q{a}a(?)求数列的首项和公比; n1
(k)(k)TTk(k,1,2,3,,,,,n)a2a,1(?)对给定的,设是首项为,公差为的等差数列.求数列的前10项之kk
和;
(i)TS,b,b,,,,,bSm(m,1)bi(?)设为数列的第项,,求,并求正整数,使得 in12nn
Sn存在且不等于零. limn,,m
(注:无穷等比数列各项的和即当时该无穷数列前n项和的极限) n,,
a,1,9a,3,,11,q,,: (?)依题意可知, 28解,,,22q,a811,,,3,2,51,q,
n,12,,(2)Tt,a,2d,2a,1,3(?)由(?)知,,所以数列的的首项为,公差, 3a,,,,122n3,,
1(2)T,即数列的前10项之和为155. S,10,2,,10,9,3,155102
i,12,,a,,,,,i,12a,1,,,,b2i,1a,i,1) ===, (?,,,,32i,1,i,1,,iiii3,,
nnS2nn,14518n,272nn,1,,,,,,,,n,,S,45,18n,27,,=lim ,,lim,,,,nmmmmn,,n,,323nnn2n,,,,
SS1nn当m=2时,=,,当m>2时,=0,所以m=2 limlimmm,,,,nn2nn
1a,1a,2(3,4,)n,{}b{}a29、 (2008广东文数)设数列满足,, 。数列aaa,,(2)12nnnnn,,123bbn,,1,(2,3,)k满足是非零整数,且对任意的正整数和自然数,都有m1n
,,,,,,11bbb。 mmmk,,1
{}b{}a(1)求数列和的通项公式; nn
cnabn,,(1,2,){}cS(2)记,求数列的前项和。 nnnnnn
12(3)n,29解析(1)由得 aaa,,aaaa,,,,()()nnn,,12nnnn,,,11233
n,122,,aa,,,10aa, 又 , 数列是首项为1公比为的等比数列, ,aa?,,,,,21nn,1,,nn,133,,
aaaaaaaaaa,,,,,,,,,,()()()() nnn12132431,
n,12,,1,,n,122n,,,8322223,,,,,,,,,,,,,,,1 , ,,,,,,,,,11,,,,,,,,2553333,,,,,,,,1,3
,,,,11bb,,,,11bb,,1223,,b,,1b,1 由 得 ,由 得 ,„ ,,,11b,,,11b,,2323,,bZb,,,0bZb,,,02233,,
当n为奇数时
1,当n为偶数时 b,,1b,1 同理可得当n为偶数时,;当n为奇数时,;因此 b,,nnn-1,
n,1,832,,当n为奇数时 nn,,,,553,,,Sccccc,,,,,, (2) nn1234cnab,,,nnnn,1832,,,当n为偶数时 nn,,,,,553,,,
当n为奇数时,
01231n,,,88888322222,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,Snn(234)1234,,n,,,,,,,,,,55555533333,,,,,,,,,,,,,,
01231n,,,41n,,,322222,,,,,,,,,, ,,,,,,,,,,,1234n,,,,,,,,,,,,5533333,,,,,,,,,,,,,,
当n为偶数时
01231n,,,88888322222,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,Snn(234)1234,,n,,,,,,,,,,55555533333,,,,,,,,,,,,,,
01231n,,,4322222n,,,,,,,,,, 1234n,,,,,,,,,,,,,,,,,,,,,,,,5533333,,,,,,,,,,,,,,
01231n,22222,,,,,,,,,,令 „„? Tn,,,,,,,,,,1234,,,,,,,,,,n33333,,,,,,,,,,
1234n2222222,,,,,,,,,,?×得: „„? Tn,,,,,,,,,,1234,,,,,,,,,,n3333333,,,,,,,,,,
12341nn,1222222,,,,,,,,,,,,?-?得: Tn,,,,,,,,1,,,,,,,,,,,,n3333333,,,,,,,,,,,,
n2,,n1,nn,,2,,223,,,,,, Tn,,,?993,,,,,,,nn33,,n,,,,,,2333,,,,,,1,3
n,93n,,,4232n,,,当n为奇数时 ,,,,553,,,S,因此 ,nn93n,,,4272n,,,,当n为偶数时 ,,,,,553,,,
1xa,1f(x),a(a,0,{a}30. (2009广东文科)已知点(1,)是函数且)的图象上一点,等比数列n3
,f(n),c(b,0){b}SSSSS的前n项和为,数列的首项为c,且前n项和满足,=+(n2). nnnnn,1nn,1
{a}{b}(1)求数列和的通项公式; nn
10001TT(2)若数列{前n项和为,问>的最小正整数n是多少? }nn2009bbnn,1
x11,,【解析】(1), 30?,fxQfa1,,,,,,,,33,,
12afcfc,,,,21,,,, ,, ,,afcc,,,,1,,,,,,21,,,,93
2 . afcfc,,,,,,,,,,32,,,,3,,,,27
42a21281c,1a又数列成等比数列, ,所以 ; ac,,,,,,,,n12a333,27
nn,1211a1,,,,*2nN,又公比,所以 ; aq,,,,,,2,,,,na3333,,,,1
n,2 QSSSSSSSS,,,,,,,,,,,,nnnnnnnn,,,,1111
b,0S,0?,,SS1又,, ; nnnn,1
2Snn,,,,,111Sn,数列构成一个首相为1公差为1的等差数列, , S,,,,nnn
22n,2bSSnnn,,,,,,,121当, ; ,,nnn,1
*?,,bn21nN,(); n
11111111(2) T,,,,,L,,,,,Knbbbbbbbb133557(21)21,,,,,,nn,,1223341,nn
1111111111111n,,,,,,,,,, ; ,,,,,,,,,1K,,,1,,,,,,,,,,2323525722121nn,,22121nn,,,,,,,,,,,,
n100010001000 由得,满足的最小正整数为112. T,T,,n,nn212009n,92009
31((本小题满分14分)
2*333a,0n,Naaaaaa,,,,,,,a已知数列满足对任意的,都有,且( ,,,,nn1212nn
aa(1)求,的值; 12
a(2)求数列a的通项公式; ,,nn
,,11S(3)设数列的前项和为,不等式对任意的正整数恒成立,求实 Sa,,log1nn,,,,nnaaa3nn,2,,
数的取值范围( a
31((2010广东文数)
(本小题主要考查数列通项、求和与不等式等知识,考查化归与转化的数学思想方法,以及抽象概括能力、
运算求解能力和创新意识)
32n,1aa,(1)解:当时,有, 11
a,1a,0由于,所以( 1n
233n,2aaaa,,,当时,有, ,,1212
a,1a,2a,0将代入上式,由于,所以( 12n
2333aaaaaa,,,,,,,(2)解:由于, ? ,,1212nn
23333aaaaaaaa,,,,,,,,,则有( ? ,,,,121121nnnn
223aaaaaaaa,,,,,,,,,?,?,得, ,,,,,,nnnn112112
2a,0aaaaa,,,,,2由于,所以( ? ,,nnnn,,1121
2aaaaa,,,,,2n?2同样有, ? ,,,,nnn,121
22aaaa,,,?,?,得( ,,nnnn11
aa,,1所以( nn,1
aa,,1aa,,1n?1a由于,即当时都有,所以数列是首项为1,公差为1的等差数列( ,,nn,121n
an,故( n
11111,,an,(3)解:由(2)知,则( ,,,n,,aannnn,,222,,,,nn,2
11111所以 S,,,,,,naaaaaaaaaa132435112,,,nnnn
11111111111111,,,,,,,,,, ,,,,,,,,,,,1,,,,,,,,,,2322423521122nnnn,,,,,,,,,,,,,
11113111,,,, ( ,,,,1,,,,,,,2212nn,,4212nn,,,,,,
1S?,?数列单调递增( SS,,,0,,nnn,1nn,,13,,,,
1所以( SS,,,,n1min3
111要使不等式对任意正整数恒成立,只要( Sa,,,,alog1nlog1,,,,naa333
10,,a01,,a?,?(
11,,aa?,即( 0,,a2
1,,所以,实数的取值范围是( 0,a,,2,,
32、(2011•广东文数)设b,0,数列{a满足a=b,a=(n?2) n}1n
(1)求数列{a}的通项公式; nn+1(4)证明:对于一切正整数n,2a?b+1( n
考点:数列递推式;数列与不等式的综合。
专题:综合题;分类讨论;转化思想。
分析
定性数据统计分析pdf销售业绩分析模板建筑结构震害分析销售进度分析表京东商城竞争战略分析
:(1)由题设形式可以看出,题设中给出了关于数列a的面的一个方程,即一个递推关系,所以应该n对此递推关系进行变形整理以发现其中所蕴含的规律,观察发现若对方程两边取倒数则可以得到一个类似
等差数列的形式,对其中
参数
转速和进给参数表a氧化沟运行参数高温蒸汽处理医疗废物pid参数自整定算法口腔医院集中消毒供应
进行讨论,分类求其通项即可( (2)由于本题中条件较少,解题思路不宜用综合法直接分析出,故求解本题可以采取分析法的思路,由
结论探究其成立的条件,再证明此条件成立,即可达到证明不等式的目的( 32解答:解:(1)?(n?2),
?(n?2),
当b=1时,(n?2),
?数列{a}是以为首项,以1为公差的等差数列, n
?=n,即a=1, n
当b,0,且b?1时,(n?2),
即数列{}是以=为首项,公比为的等比数列, ?=×=,即a=, n
?数列{a}的通项公式是 n
(2)证明:当b=1时,不等式显然成立
n+1当b,0,且b?1时,a=,要证对于一切正整数n,2a?b+1,只需证nn
n+12×?b+1,即证
?
=
,,n+1n1n2=(b+1)×(b+b+…+b+1) ,,,2n2n1n+2n+1n1n2=(b+b+…+b+b)+(b+b+…+b+1)
,nnn12=b[(b+b+…+b+b)+(++…+)]
nn?b(2+2+…+2)=2nb
所以不等式成立,
n+1综上所述,对于一切正整数n,有2a?b+1, n
点评:本题考点是数列的递推式,考查根据数列的递推公式求数列的通项,研究数列的性质的能力,本题中递推关系的形式适合用取倒数法将所给的递推关系转化为有规律的形式,两边取倒数,条件许可的情况下,使用此技巧可以使得解题思路呈现出来(数列中有请多成熟的规律,做题时要注意积累这些小技巧,在合适的情况下利用相关的技巧,可以简化做题(在(2)的证明中,采取了分析法的来探究解题的思路,通过本题希望能进一步熟悉分析法证明问题的技巧(
33((2012广东文数)(本小题满分14分)
2*sasTTSnnN,,,2,设数列的前项和,数列的前项和为,满足( nn,,,,,,nnnnnn
a(1) 求的值; 1
a(2) 求数列的通项公式( ,,n
33. 解:(1):
2a,2a,1„„„„„„„„„„„„„„„„„„3分 11
a,1„„„„„„„„„„„„„„„„„„„„„„5分 1
(2)
2? T,2S,n???nn2?„„„„„„„„„„6分 T,2S,(n,1)???n,n,11
?-?得:
S,2a,2n,1 „„„„„„ ?„„„„„„„„„7分 nn
在向后类推一次 S,2a,2(n,1),1„„„ ?„„„„„„„„„„8分 n,1n,1
?-?得:
a,2a,2a,2„„„„„„„„„„„„„„„„9分 nnn,1a,2a,2„„„„„„„„„„„„„„„„„„„10分 nn,1
a,2,2(a,2)„„„„„„„„„„„„„„„„„12分 nn,1{a,2}是以首项为a,2,3,公比为2的数列„„„„13分 n1
n,1?a,2,3,2 n
n,1?a,3,2,2„„„„„„„„„„„„„„„„„„14分 n