- 问题详情:如果{an}为递增数列,则{an}的通项公式可以为( ).A.an=-2n+3 B.an=-n2-3n+1C.an= D.an=1+log2n【回答】D知识点:数列题型:选择题...
- 20134
- 问题详情:已知数列{an}中,a1=2,n(an+1﹣an)=an+1,n∈N*,若对于任意的a∈[﹣2,2],n∈N*,不等式<2t2+at﹣1恒成立,则实数t的取值范围为()A.(﹣∞,﹣2]∪[2,+∞) B.(﹣∞,﹣2]∪[1,+∞) C.(﹣∞,﹣1]∪[2,+∞) D.[﹣2,2]【回答】A.知识点:数列题型:选择题...
- 23616
- 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:已知数列{an}满足an+1=an+2n+1,a1=1,则a5= .【回答】25.【考点】8H:数列递推式.【分析】an+1=an+2n+1,可得an﹣an﹣1=2(n﹣1)+1.(n≥2).利用累加求和实数即可得出.【解答】解:∵an+1=an+2n+1,∴an﹣an﹣1=2(n﹣1)+1.(n≥2).∴an=(an﹣an﹣1)+(an﹣1﹣an﹣2)+…+(a3﹣a2)+(a2﹣a1)+a1=2(n﹣1)+1+2(n﹣2)+1+…+2+1+1=2×+n=n2.则a5...
- 15115
- 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:若数列{an}的前n项和为Sn=an+,则数列{an}的通项公式是an=______.【回答】;【详解】试题分析:解:当n=1时,a1=S1=a1+,解得a1=1,当n≥2时,an=Sn-Sn-1=()-()=-整理可得an=−an−1,即=-2,故数列{an}是以1为首项,-2为公比的等比数列,故an=1×(-2)n-1=(-2)n-1故*为(-2)n-1.考点:等比数列的通项公...
- 27488
- 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: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:已知数列{an}中,a1=1,an=2an﹣1+2n(n≥2),则an= .【回答】(2n﹣1)•2n﹣1.【考点】8H:数列递推式.【分析】an=2an﹣1+2n(n≥2),可得﹣=1,利用等差数列的通项公式即可得出.【解答】解:∵an=2an﹣1+2n(n≥2),∴﹣=1,可得数列是等差数列,公差为1,首项为.∴==,解得an=(2n﹣1)•2n﹣1.n=1时也成立.∴an=(2n﹣1)•2n﹣1.故*...
- 24835
- 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:在数列{an}中,a1=,an+1=an+(n∈N*),则a2019的值为. 【回答】1解析因为an+1=an+(n∈N*),所以an+1-an=,a2-a1=1-,a3-a2=,……a2019-a2018=,累加,可得a2019-a1=1-,a2019-=1-,所以a2019=1.知识点:数列题型:填空题...
- 15668
- 1、aneyeforaneye;titfortat2、Retributivejusticedemandsaneyeforaneye.3、Thedeathpenaltyformurderworkontheprincipleofaneyeforaneye.4、Thiswastheruleofaneyeforaneyeandatoothforatooth.5、EventheBiblesaysaneyeforaneyeandatoothforatooth.6、Kevindecidestogi...
- 13340
- 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:若数列{an}的前n项和Sn=an+,则数列{an}的通项公式是an=.【回答】由S1=a1+=a1,解得a1=1,又Sn=an+,得Sn-1=an-1+(n≥2),所以Sn-Sn-1=an-an-1=an,得=-2.所以数列{an}是首项为1,公比为-2的等比数列.故数列的通项公式an=(-2)n-1.*:(-2)n-1知识点:数列题型:填空题...
- 15424
- 问题详情:An_______musttake_______lessons.A.actor;acting B.actor;actC.acting;actor ...
- 11243
- 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: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
- 问题详情:数列{an}中,a1=2,且an+1=an-1,则a5的值为________.【回答】-知识点:数列题型:填空题...
- 19648
- 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:设无穷数列{an}满足:n∈Ν,an<an+1,an∈N.记bn=aan,cn=aan+1(n∈N*).(1)若bn=3n(n∈N*),求*:a1=2,并求c1的值;(2)若{cn}是公差为1的等差数列,问{an}是否为等差数列,*你的结论.【回答】解:(1)因为an∈N,所以若a1=1,则b1=aa1=a1=3矛盾,若a1≥3=aa1,可得1≥a1≥3矛盾,所以a1=2.于是a2=aa1=3,从而c1=aa1+1...
- 10310
- 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