- 问题详情:JoannewasstuckinatrafficjamincentralBirminghamat5:30.andat6:30shewasexpectedtobechairingameetingofthetennisclub.Atlast,thetrafficwasmoving.SheswungquicklyracingtoherhouseAssheopenedthedoor,shenearlytrippedoverSheba. “Hey,Sheba,”shesaid,“I...
- 18015
- 问题详情:若数列{an}的前n项和Sn=an+,则{an}的通项公式是an=________.【回答】(-2)n-1知识点:数列题型:填空题...
- 8487
- 问题详情:如图,∠MAN=30°,点O为边AN上一点,以O为圆心,4为半径作⊙O交AN于D,E两点.(1)当⊙O与AM相切时,求AD的长;(2)如果AD=2,那么AM与⊙O又会有怎样的位置关系?并说明理由.【回答】解:(1)设AM与⊙O相切于点B,并连接OB,则OB⊥AB;在△AOB中,∠A=30°,则AO=2OB=8,所以AD=AO﹣OD,即AD=4.(2)AM与⊙O相交,理由...
- 9459
- 问题详情:设Sn是数列{an}的前n项和,an>0,且4Sn=an(an+2).(1)求数列{an}的通项公式;(2)设bn=,Tn=b1+b2+…+bn,求*:Tn<.【回答】(1)解4Sn=an(an+2),①当n=1时,4a1=+2a1,即a1=2.当n≥2时,4Sn-1=an-1(an-1+2). ②由①-②得4an=+2an-2an-1,即2(an+an-1)=(an+an-1)·(an-an-1...
- 7699
- 问题详情:设数列{an}满足an+1=an+2,a1=4.(1)求*:{an-3}是等比数列,并求an;(2)求数列{an}的前n项和Tn.【回答】解(1)∵an+1=an+2,a1=4,∴an+1-3=(an-3).故{an-3}是首项为1,公比为的等比数列.∴an=3+n-1.(2)an=3+n-1,故Tn=3n+0+1+…+n-1=3n+=3n+1-n.知识点:数列题型:解答题...
- 18049
- 问题详情:设等比数列{an}的前n项和为Sn,已知a2=6,6a1+a3=30,求an和Sn【回答】解:设的公比为q,由题设得 …………4分解得 ...
- 4171
- 问题详情: 在数列{an}中,若a1=1,an+1=an+2(n≥1),则该数列的通项an=________.【回答】an=2n-1解析:由已知{an}为等差数列,d=an+1-an=2,∴an=2n-1.知识点:数列题型:填空题...
- 18674
- 问题详情: 已知a1=1,an=n(an+1-an)(n∈N*),则数列{an}的通项公式是________.【回答】an=n知识点:数列题型:填空题...
- 32465
- 问题详情:计算:an+2·an+1·an·a.【回答】解:原式=an+2+n+1+n+1=a3n+4.知识点:整式的乘法题型:计算题...
- 23000
- 问题详情:数列{an}满足a1=a2=1,an+an+1+an+2=cos(n∈N*),若数列{an}的前n项和为Sn,则S2013的值为()A.2013 B.671 C.-671 D.-【回答】D知识点:数列题型:选择题...
- 20211
- 问题详情:已知在等差数列{an}中,a7+a9=16,a4=1,则a12的值是A.15B.30 C.31D.64【回答】解析由等差数列的*质得a7+a9=a4+a12,因为a7+a9=16,a4=1,所以a12=15.故选A.*A知识点:数列题型:选择题...
- 27770
- 问题详情:如图,已知∠MAN=30°,O为边AN上一点,以O为圆心,2为半径作⊙O,交AN于D,E两点,设AD=x.当x为何值时,⊙O与AM相交于B,C两点,且∠BOC=90°?【回答】解:过点O作OF⊥BC于点F.∵∠BOC=90°,OB=OC=2,∴∠OBC=45°,BC==2.∵OF⊥BC,∴BF=BC=,∠BOF=45°.∴∠OBF=∠BOF.∴OF=BF=.∵∠MAN=30°,∴OA=2OF=2.∴AD=2-2,即...
- 15631
- 问题详情:等比数列{an}中,a1+a2=30,a3+a4=60,则a7+a8=( )(A)120 (B)180 (C)240 (D)270【回答】C知识点:数列题型:选择题...
- 7770
- 问题详情:已知am=5,an=6,则am+n的值为()A.11 B.30 C. D.【回答】B【考点】同底数幂的乘法.【分析】根据同底数幂的乘法法则:同底数幂相乘,底数不变,指数相加,进行计算即可.【解答】解:am+n=am×an=30.故选B.知识点:整式的乘法题型:选择题...
- 32155
- 问题详情:已知数列{an}中,a1=1,an+1=an+3,若an=2017,则n=()A.667 B.668 C.669 D.673【回答】D知识点:数列题型:选择题...
- 29555
- 问题详情:在等差数列{an}中,S10=10,S20=30,则S30= .【回答】60【解析】根据等差数列的*质,利用S10,S20-S10,S30-S20成等差数列列式求解即可.【详解】【点睛】本题主要考查了等差数列*质的运用,属于基础题. 知识点:数列题型:填空题...
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- 问题详情:已知等差数列{an}中,a1+a3+a9=20,则4a5-a7=()A.20 B.30 C.40 D.50【回答】A 知识点:数列题型:选择题...
- 22214
- 1、aneyeforaneye;titfortat2、Retributivejusticedemandsaneyeforaneye.3、Thedeathpenaltyformurderworkontheprincipleofaneyeforaneye.4、Thiswastheruleofaneyeforaneyeandatoothforatooth.5、EventheBiblesaysaneyeforaneyeandatoothforatooth.6、Kevindecidestogi...
- 13340
- 问题详情:在数列{an}中,a1=1,a2=2,若an+2=2an+1-an+2,则an等于()A. B.n3-5n2+9n-4C.n2-2n+2 D.2n2-5n+4【回答】C命题立意:本题考查等差数列的定义与通项公式、累加法求数列的通项公式,...
- 27526
- 问题详情:数列{an}满足a1=1,a2=2,an+2=2an+1﹣an+2.(Ⅰ)设bn=an+1﹣an,*{bn}是等差数列;(Ⅱ)求{an}的通项公式.【回答】解答:解:(Ⅰ)由an+2=2an+1﹣an+2得,an+2﹣an+1=an+1﹣an+2,由bn=an+1﹣an得,bn+1=bn+2,即bn+1﹣bn=2,又b1=a2﹣a1=1,所以{bn}是首项为1,公差为2的等差数列.(Ⅱ)由(Ⅰ)得,bn=1+2(n﹣1)=2n﹣1,由bn=an+1﹣an得,a...
- 20406
- 问题详情:已知等差数列{an}中,a6+a10=16,a4=1,则a12的值是A.15 B.30 C.31 D.64【回答】A知识点:数列题型:选择题...
- 14038
- 问题详情:已知数列{an}满足an+1=an﹣an﹣1(n≥2),a1=m,a2=n,Sn为数列{an}的前n项和,则S2017的值为()A.2017n﹣m B.n﹣2017m C.m D.n【回答】C【考点】数列递推式.【分析】an+1=an﹣an﹣1(n≥2),a1=m,a2=n,可得an+6=an.即可得出.【解答】解:∵an+1=an﹣an﹣1(n≥2),a1=m,a2=n,∴a3=n﹣m,a4=﹣m,a5=﹣n,a6=...
- 11959
- 问题详情:在数列{an}中,若a1=1,an+1=an+2n,则an等于()A.2n-1 B.2n+1-3 C.2n-1 D.2n-1-1【回答】A知识点:数列题型:选择题...
- 6839
- 问题详情:数列{an}中,已知a1=1,a2=2,an+2=an+1-an(n∈N*),则a2017=()A.1 B.-1 C.-2 D.2【回答】A 知识点:数列题型:选择题...
- 27666
- 问题详情:在数列{an}和{bn}中,an+1=an+bn+,bn+1=an+bn-,a1=1,b1=1.设cn=+,则数列{cn}的前2018项和为________.【回答】4036解析:由已知an+1=an+bn+,bn+1=an+bn-,得an+1+bn+1=2(an+bn),所以=2,所以数列{an+bn}是首项为2,公比为2的等比数列,即an+bn=2n,将an+1=an+bn+,bn+1=an+bn-相乘,得=2,所以数列{anbn}是首项为1,公比为2的等比数列...
- 18230