第45练数列的递推关系及通项[基础保分练]1.已知数列{an}的通项公式为an=则a20a15=________.2.在数列{an}中,a1=2,an+1=-2an+3,则数列{an}的通项公式an=________.3.已知数列{an}满足a1=0,an+1=(n=1,2,3,…),则a2019=________.4.已知数列{an}的前n项和为Sn,且Sn=2an-2,则a2019=________.5.由a1=1,an+1=给出的数列{an}的第34项是________.6.正项数列{an}中,满足a1=1,a2=,=(n∈N*),那么an=________.7.已知数列{an}的首项a1=2,且(n+1)an=nan+1,则a5=________.8.数列{an}中,a1=1,且an+1=an+2n,则a9=________.9.数列{an}中,若a1=1,an+1=an,则an=________.10.已知数列{an}的前n项和为Sn,a1=1,2Sn=(n+1)an,则an=________.[能力提升练]1.已知数列{an}的通项为an=,当an取得最小值时,n的值为________.2.已知数列{an},若a1=2,an+1+an=2n-1,则a2019=________.3.设各项均为正数的数列{an}的前n项和为Sn,且Sn满足2S-(3n2-n-4)Sn-2(3n2-n)=0,n∈N*,则数列{an}的通项公式是________.4.已知数列{an}中,a1=2,n(an+1-an)=an+1,n∈N*.若对于任意的t∈[0,1],n∈N*,不等式<-2t2-(a+1)t+a2-a+3恒成立,则实数a的取值范围为______________.5.已知数列{an}满足an=若对任意n∈N*,都有an>an+1,则实数a的取值范围是______________.6.设数列{an}的前n项和为Sn,且Sn=n2-n+1,正项等比数列{bn}的前n项和为Tn,且b2=a2,b4=a5,数列{cn}中,c1=a1,且cn=cn+1-Tn,则{cn}的通项公式为______________.答案精析基础保分练1.1652.(-2)n-1+13.4.220195.6.解析由已知2=·,∴数列是等比数列,又=1,=2,∴q=2,∴=2n-1,∴an=.7.10解析由题意得an+1=an,∴a2=a1=4,a3=a2=6,a4=a3=8,a5=a4=10.8.511解析由题意可得an+1-an=2n,则a9=a1+(a2-a1)+(a3-a2)+…+(a9-a8)=1+21+22+…+28=29-1=511.9.解析∵a1=1,an+1=an,则(n+1)an+1=nan=a1=1,∴an=.10.n解析由2Sn=(n+1)an知,①当n≥2时,2Sn-1=nan-1,②①-②,得2an=(n+1)an-nan-1,∴(n-1)an=nan-1,∴当n≥2时,=,∴==1,∴an=n.能力提升练1.152.20203.an=3n-24.(-∞,-1]∪[3,+∞)解析根据题意,数列{an}中,n(an+1-an)=an+1,∴nan+1-(n+1)an=1,∴-=-,∴=++…++a1,=++…++2=3-<3,∵<-2t2-(a+1)t+a2-a+3恒成立,∴3≤-2t2-(a+1)t+a2-a+3.∴2t2+(a+1)t-a2+a≤0,在t∈[0,1]上恒成立,设f(t)=2t2+(a+1)t-a2+a,t∈[0,1],∴即解得a≤-1或a≥3.5.6.cn=2n-n解析∵Sn=n2-n+1,∴令n=1,a1=1,an=Sn-Sn-1=2(n-1)(n≥2),经检验a1=1不符合上式,∴an=又∵数列{bn}为等比数列,b2=a2=2,b4=a5=8,∴=q2=4,又数列{bn}为正项等比数列,∴q=2,∴b1=1,∴bn=2n-1.Tn==2n-1,∵c2-c1=21-1,c3-c2=22-1,…,cn-cn-1=2n-1-1,以上各式相加得cn-c1=-(n-1),c1=a1=1,∴cn-1=2n-n-1,∴cn=2n-n.