2020-2021学年新教材数学人教A版选择性必修第二册课后提升训练：4-1　第2课时　数列的递推公式 WORD版含解析.docx

1、第四章数列4.1数列的概念第2课时数列的递推公式课后篇巩固提升基础达标练1.数列12,14,18,116,的递推公式可以是()A.an=12n+1(nN*)B.an=12n(nN*)C.an+1=12an(nN*)D.an+1=2an(nN*)解析数列从第2项起,后一项是前一项的12,故递推公式为an+1=12an(nN*).答案C2.(多选)符合递推关系式an=2an-1的数列是()A.1,2,3,4,B.1,2,2,22,C.2,2,22,4,D.0,2,2,22,解析B与C中从第2项起,后一项是前一项的2倍,符合递推公式an=2an-1.答案BC3.在数列an中,an+1=an+2-an

2、,a1=2,a2=5,则a5=()A.-3B.-11C.-5D.19解析由an+1=an+2-an,得an+2=an+an+1,则a3=a1+a2=7,a4=a2+a3=12,a5=a3+a4=19.答案D4.已知数列an的前n项和为Sn,且Sn=2an-2,则a2=()A.4B.2C.1D.-2解析Sn=2an-2a1=S1=2a1-2a1=2a1+a2=S2=2a2-2a2=4.故选A.答案A5.已知a1=1,an=n(an+1-an)(nN*),则数列an的通项公式是()A.2n-1B.n+1nn-1C.n2D.n解析法一:构造法.由已知整理,得(n+1)an=nan+1,an+1n+1

3、=ann,数列ann是常数列,且ann=a11=1,an=n.法二:累乘法.当n2时,anan-1=nn-1,an-1an-2=n-1n-2,a3a2=32,a2a1=21,两边分别相乘,得ana1=n.a1=1,an=n.答案D6.在数列an中,若a1=2,an+1=an+n-1,则a4=.解析a2=a1+1-1=2,a3=a2+2-1=3,a4=a3+3-1=5.答案57.(2020潍坊高三检测)已知数列an的前n项和Sn=n2-9n,第k项满足5ak8,则k=.解析an=S1(n=1),Sn-Sn-1(n2),an=2n-10.由52k-108,得7.5k9,k=8.答案88.若数列an

4、满足an+1=2an-1,且a8=16,则a6=.解析an+1=2an-1,a8=2a7-1=16,解得a7=172,又a7=2a6-1=172,解得a6=194.答案1949.在数列an中,a1=2,an+1=an+ln1+1n,求an.解由题意,得an+1-an=lnn+1n,an-an-1=lnnn-1(n2),an-1-an-2=lnn-1n-2,a2-a1=ln21,当n2时,an-a1=lnnn-1n-1n-221=lnn,an=2+lnn(n2).当n=1时,a1=2+ln1=2,符合上式,an=2+lnn(nN*).10.已知各项均不为0的数列an满足a1=12,anan-1=

5、an-1-an(n2,nN*),求数列an的通项公式.解anan-1=an-1-an,且各项均不为0,1an-1an-1=1.当n2时,1an=1a1+1a2-1a1+1a3-1a2+1an-1an-1=2+1+1+1(n-1)个1=n+1.1an=n+1,当n2时,an=1n+1.a1=12也符合上式,an=1n+1.能力提升练1.已知数列an,a1=2,a2=1,an+2=3an+1-an,则a6+a4-3a5的值为()A.3B.-2C.-1D.0解析an+2=3an+1-an,an+2+an=3an+1.令n=4,得a6+a4=3a5,a6+a4-3a5=0.答案D2.已知数列an,an

6、+1=11-an,a1=3,则a2 019=()A.23B.3C.-12D.32解析由题意,可知:a1=3,a2=11-a1=11-3=-12,a3=11-a2=11+12=23,a4=11-a3=11-23=3,a5=11-a4=11-3=-12,数列an是一个以3为最小正周期的周期数列.20193=673,a2019=a3=23.答案A3.已知数列an,a1=1,ln an+1-ln an=1,则数列an的通项公式是()A.an=nB.an=1nC.an=en-1D.an=1en-1解析lnan+1-lnan=1,lnan+1an=1.an+1an=e.由累乘法可得an=en-1.答案C4

7、.在数列an中,a1=1,an+1=an2-1,则此数列的前4项和为.解析a1=1,an+1=an2-1,a2=12-1=0,a3=02-1=-1,a4=(-1)2-1=0,故前4项和a1+a2+a3+a4=0.答案05.(2019浙江期末)已知数列an满足:a1=2,an=an+1an-1(nN*).设Sn为数列an的前n项和,则a2=,S2 019=.解析由题意知a2=a1+1a1-1=2+12-1=3,又a3=3+13-1=2=a1,数列an是周期数列,周期为2,S2019=(2+3)1009+2=5047.答案35 0476.已知在数列an中,an+1=2anan+2对任意正自然数n都

8、成立,且a7=12,则a5=.解析由已知a7=2a6a6+2=12,解得a6=23.又因为a6=2a5a5+2=23,解得a5=1.答案17.在数列an中,已知a1=2,a2=3,an+2=3an+1-2an(n1),写出此数列的前6项.解a1=2,a2=3,a3=3a2-2a1=33-22=5,a4=3a3-2a2=35-23=9,a5=3a4-2a3=39-25=17,a6=3a5-2a4=317-29=33.8.已知数列an满足an+1=2an,0an12,2an-1,12an1.若a1=67,试求a2 018+a2 019.解a1=6712,1,a2=2a1-1=5712,1,a3=2a2-1=370,12,a4=2a3=67.数列an是周期数列,且周期为3.a2018+a2019=a6723+2+a6733=a2+a3=57+37=87.素养培优练已知数列a1=1,a2,a3,an(nN*)的法则如下:若an为自然数,则an+1=an-2,否则an+1=an+3,则a6=.解析a1=1是自然数,a2=a1-2=1-2=-1.a2=-1不是自然数,a3=a2+3=-1+3=2.a3=2是自然数,a4=a3-2=2-2=0.a4=0是自然数,a5=a4-2=0-2=-2.a5=-2不是自然数,a6=a5+3=-2+3=1.答案1