2013年高考文科数学数列解答题
1.[2014?北京卷] 已知{a}是等差数列,满足a,3,a,12,数列{b}满足b,4,b,n14n14
20,且{b,a}为等比数列( nn
(1)求数列{a}和{b}的通项公式; (2)求数列{b}的前n项和( nnn
,aa12,341解:(1)设等差数列{a}的公差为d,由题意得d,,a,(n,,,3.所以ann133
,ab20,124431)d,3n(设等比数列{b,a}的公比为q,由题意得q,,,8,解得q,2. nnb,a4,311,,,n1n1n1所以b,a,(b,a)q,2.从而b,3n,2(n,1,2,…)( nn11n
3,,n1n1(2)由(1)知b,3n,2(n,1,2,…)(数列{3n}的前n项和为n(n,1),数列{2}的n2
n1,23nn前n项和为1×,1,所以,数列{b}的前n项和为n(n,1),2,1. ,2n21,2
2(,[2014?福建卷] 在等比数列{a}中,a,3,a,81. n25
(1)求a; (2)设b,loga,求数列{b}的前n项和S. nn3nnn
,,aq,3,a,1,11,,,n1,,解:(1)设{a}的公比为q,依题意得因此,a,3. 解得nn4,,aq,81,q,3.,,1
2,b)n,nn(b1n,. (2)因为b,loga,n,1,所以数列{b}的前n项和S,n3nnn22
aa,11. ,2013年高考重庆卷,文,,设数列满足:,,. aa,3nN,,,n1nn,1,
a(?)求的通项公式及前项和; nS,,nn
b(?)已知是等差数列,为前项和,且,,求. baaa,,,Tnba,T,,n3123n1220
2.,2013年高考四川卷,文,,在等比数列中,,且为和的等差中3a{}aaa,,22aa1n2123项,求数列的首项、公比及前项和. {}ann
2aq,a,2a解:设的公比为q.由已知可得 ,, 4aq,3a,aq,,11n111
2a(q,1),2q,3q,1所以,,解得 或 , q,4q,3,01
n3,1a,1a(q,1),2q,3q,1由于.因此不合题意,应舍去,故公比,首项.所以, S,11n2
a3. ,2013年高考课标?卷,文,,已知等差数列的公差不为零,a=25,且a,a,a成等111113,,n
a比数列. (?)求的通项公式; (?)求. aaaa,,,,?,,n14732n,
4.,2013高考浙江,在公差为d的等差数列{a}中,已知a=10,且a,2a+2,5a成等比数列. n1123
(?)求d,a; (?) 若d<0,求|a|+|a|+|a|++|a| . n123n
222解:(?)由已知得到: (22)54(1)50(2)(11)25(5)aaaadaddd,,,,,,,,,,,21311
dd,,,41,,22; ,,,,,,,,,,1212212525340ddddd或,,anan,,,,4611nn,,
nn(21),,(?) ; ,(111),,n,2,123n2||||||||aaaa,,,,, ,nn,,21220,,(12)n,,,2
a5.,2013年高考大纲卷,文,,等差数列中, aaa,,4,2,,,n7199
1bbnS,求数列的前项和a,.(I)求的通项公式; (II)设 ,,,,nnnnnan
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答案
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】(?)设等差数列的公差为d,则 aand,,,(1){}an1n
a,4ad,,64,,1n,171因为,所以. 解得,.所以. ad,,1,a,,,1naa,2adad,,,182(8)2219911,,
12222222222nb,,,,(?), 所以. S,,,,,,,,()()()?nnnannnn,,(1)1122311nnn,,n
26. ,2013年高考江西卷,文,,正项数列{a}满足. anan,,,,(21)20nnn
1b,(1)求数列{a}的通项公式a; (2)令,求数列{b}的前n项和T. nnnnn(1)na,n
2【答案】解:(1)由aa,,,,(21)20nn得(a-2n)(a+1)=0则. a,2nnnnnn
11111b,,,,()(2)由(1)知,故 a,2nnnnannnn,,,(1)(1)(2)2(1)n
11111111n ?,,,,,,,,,,T(1...)(1)n222312122nnnn,,,
7. ,2013年高考课标?卷,文,,已知等差数列的前n项和满足,. {}aSS,0S,,5nn35
1(?)求的通项公式; (?)求数列的前n项和. {}a{}naa2121nn,,
nn(1),n【答案】(1)设{a}的公差为d,则S=. nad,1n2
330,ad,,,1解得ad,,,1,1.,1 由已知可得故的通项公式为aan=2-.,,5105,ad,,,nn1,
11111,,,(), (2)由(I)知 aannnn,,,,(32)(12)223212121nn,,
,,11111111nn. 从而数列的前项和为(-+-++)?,,,,aa2-1113232112nnn,,,2121nn,,,,
a8.,2013年高考广东卷,文,,设各项均为正数的数列的前项和为,满足Sn,,nn
2,且构成等比数列. 441,,SannN,,,,aaa,,nn2514,1
aa,,45a(1) 证明:; (2) 求数列的通项公式; ,,21n
1111,,,,?(3) 证明:对一切正整数,有. naaaaaa212231nn,
22?aaa,?,,045n,1时,, 【答案】(1)当45,45aaaa,,,,n211221
222n,24411San,,,,(2)当时,, 4444aSSaa,,,,,,,nn,nnnnn,,111
222aaaa,,,,,442n,2ad,2, 当时是公差的等差数列. ?aaa,?,,02?,,,,1nnnn,nnnn,1
22aaa,,,,824构成等比数列,,,解得, ?,,aaa?aaa,,a,3,,,,222251425214
2a由(1)可知, 是首项,公差45=4,1aaa,,?,a,1?aa,,,,312?,,n211121
d,2a的等差数列. 数列的通项公式为. an,,21?,,nn
1111111,,,,,,,,??(3) aaaaaann,,,,,1335572121,,,,12231nn,
11111111,,,,,,,,,, ,,,,,,,,,1,,,,,,,,,,2335572121nn,,,,,,,,,,,,
111,,,,,,1.,,2212n,,,
9. ,2013年高考湖南,文,,设为数列{}的前项和,已知, Saa,0nn1
,2,N(?)求,,并求{}的通项公式;(?)求{}的前项和. n,aana,a,S,Sanan11n12nn【答案】解: (?) ?S,a.?当n,1时,2a,a,S,S,a,0,a,1.11111111
2a,a2a,an1n,11- 当n,1时,a,s,s,,,2a,2a,a,2annn,1nn,1nn,1
SS11
n,1 ,{a}时首项为a,1公比为q,2的等比数列,a,2,n,N*.n1n
(?) 设T,1,a,2,a,3,a,?,n,a,qT,1,qa,2,qa,3,qa,?,n,qan123nn123n
n1,qnn(1,q)T,a,a,a,,?a,na,a,na,2,1,n,2 ,qT,1,a,2,a,3,a,?,n,an123nn,11n,1n234n,11,q
n. ,T,(n,1),2,1,n,N*n
,,n10.,2013年高考山东卷,文,,设等差数列a的前项和为S,且S,4S,a,2a,1 n2nnn42
bbb1*n12(?)求,,,,,,na的通项(?)设b满足 ,求b的前项和T ,,,,,, 1,nNnnnnn2aaan12