构造辅助数列求通项公式(1)同步练习 高二上学期数学苏教版（2019）选择性必修第一册

培优练❹构造辅助数列求通项公式(1)A级基础达标练1.在数列{an}中,a1=1,an+1-an=2n-1,则a1+a2+…+a9=()A.958B.967C.977D.9972.已知数列{an}满足an=an-12an-1+1(n≥2,n∈A.2n B.n2 C.3n-1 D.3.已知数列{an}满足a1=2,an+1-an+n=0,则a100=.

4.已知数列{an}满足a1=1,an+1=an+2n-1,则an=.

5.在数列{an}中,a1=1,当n≥2时,an=2nan-1,则数列{an}的通项公式为.

6.已知数列{an}满足a1=-1,an-an+17.在数列{an}中,已知a1=1p,an+1=a(1)若p=1,求数列{an}的通项公式;(2)记bn=nan,若在数列{bn}中,bn≤b4,求实数p的取值范围.B级能力提升练8.已知数列{an}的前n项和为Sn,a2=6,Sn=(n+1)A.an=3n B.an=3nC.an=n+4 D.an=n2+29.已知数列{an}满足a1=1,a2=2,an+2-2an+1+an=12,则a100A.50492 B.252510.在数列{an}中,a1=1,an+1=an3aA.34103 B.100 C.110011.在数列{an}中,a1=0,an+1-an=1n+n+1且a12.已知数列{an}满足a1=2,且a12+a23+a313.已知数列{an}满足a1=23,(2n+2-1)an+1=(2n+1-2)an,则{an}的通项公式为14.在正项数列{an}中,a1=1,a2=10,anan-1=a15.已知数列{an}满足an+1=an2n+1·an+1.若a1=12,则∑i=1C级拓展探究练16.已知数列{an}满足a1=1,若1an+1-1an=4n(n∈A.34n-1 B.43参考答案1.C2.A3.-49484.n2-2n+25.an=2n2+n7.解(1)由an+1=annan+1,得1an+1-1an=n,由累加法得1a2-1a1+1a3-1a2+…+1an+1-1an=1+2+…+n=n(n(2)由(1)知an=2n2-n+2p,bn=2nn2-n+2p,则b4=46+p.因为bn≤b4,所以2nn2-n+2p≤46+p,化简得(n-4)p≤2n(n-4).当n<4时,p≥2n,即p≥(2n)8.A9.C10.C11.10012.an=n+1解析依题意,数列{an}满足a1=2,且a12+a23+a34+…+an-1n=an-2①.当n=2时,a12=a2-2,a2=3,a12+a23+a34+…+an-1n+ann+1=an+1-2②,②-①,得ann+1=an+1-an,an+113.an=2n(2n-1)(2n+1-1)解析2n+1-22n+2an-2an-3·…·a2a1=2·2n-1-12n+1-1·2·2n-2-12n-1·14.102-(12)n-2解析对anan-1=an-1an-2两边同时取常用对数,可得lgan-lgan-1=12(lgan-1-lgan-2).令bn=lgan+1-lgan,则b1=lga2-lga1=1,bn-1=12bn-2(n=3,4,5,…),所以bn=(12)n-1(n=1,2,3,…),所以lgan+1-lgan=因为1+12+14+1-12n-11-1102[1-(12)n-1],则an=102-(12)n-2,a15.260412解析由an+1an2n+1·an+1取倒数,得1an+1=1an+2n+1,即1an+1-1an=2n+1,则当n≥2时,1an=1a1+1a2-1a1+1a3-1a2+…+1an-1an-1=1a1+22+23+…+2n=2n+1-4+1a1;当n=1时,上式也成立,则1an=2n+1-4+1a1.当a1=12时,116.A