1、第五章数列5.1数列基础5.1.2数列中的递推课后篇巩固提升基础达标练1.数列1,3,6,10,15,的递推公式是()A.a1=1,an+1=an+n,nN+B.a1=1,an=an-1+n,nN+,n2C.a1=1,an+1=an+(n+1),nN+,n2D.a1=1,an=an-1+(n-1),nN+,n2解析由题可知a1=1,an-an-1=n(n2).答案B2.已知数列an中的首项a1=1,且满足an+1=12an+12n,此数列的第3项是()A.1B.12C.34D.58解析a1=1,a2=12a1+12=1,a3=12a2+122=34.答案C3.已知数列an中,a1=1,an+1
2、=2an+1,则数列an的一个通项公式为()A.an=nB.an=n+1C.an=2nD.an=2n-1解析由题可知a1=1,a2=3,a3=7,a4=15,经验证,选D.答案D4.(2020黑龙江大庆中学高一月考)已知数列an的前n项和为Sn(nN+),且Sn=n2+.若数列an为递增数列,则实数的取值范围为()A.(-,1)B.(-,2)C.(-,3)D.(-,4)解析当n=1时,a1=S1=1+;当n2时,an=Sn-Sn-1=n2+-(n-1)2-=2n-1,因为an+1-an=20(n2),所以当n2时,数列an为递增数列.若数列an为递增数列,只需a2a1,所以31+,即0,an0
3、,解得n6或n6(nN+)时,an0.令n2-n-300,解得0n6,当0n6(nN+)时,an0.(3)Sn存在最小值,不存在最大值.由an=n2-n-30=n-122-1214,知an是递增数列,且a1a2a5a6=0a7a8a91,则数列an为递增数列B.若数列an为递增数列,则a1C.存在实数a,使数列an为常数数列D.存在实数a,使|an+1|2恒成立解析对于A选项,若a1,则an+1-an=an2+an-2=an+122-941+122-94=0,an+1an,即数列an为递增数列,则A正确;对于B选项,若数列an为递增数列,则an+1-an=an2+an-2=an+122-940
4、,an+1232,即an1,a1,则B错;对于C选项,要使数列an为常数数列,则an+1-an=an2+an-2=(an-1)(an+2)=0,an=1,或an=-2,即存在实数a=1或a=-2,使数列an为常数数列,则C正确;对于D选项,由C选项可得,当a=1时,数列an为常数数列,即|an+1|=|1+1|=2,则存在实数a=1,使|an+1|2恒成立,则D正确.答案B3.(2019浙江宁波高三月考)已知数列an,满足a1=3,an+1an=an+2(nN+),则使an42 020成立的最小正整数n为()A.10B.11C.12D.13解析因为an+1an=an+2,即an+1=an2+2
5、an,所以an+1+1=an2+2an+1=(an+1)2,则a2+1=(a1+1)2,a3+1=(a2+1)2=(a1+1)22,an+1=(an-1+1)2=(a1+1)2n-1,所以an+1=42n-1,即an=42n-1-1,因为an42020,即42n-1-142020,又nN+,所以n12,故选C.答案C4.数列an中,a1=7,a9=8,且(n-1)an=a1+a2+an-1(n3),则a2等于.解析由(n-1)an=a1+a2+an-1(n3),得nan+1=a1+a2+an,两式相减,得nan+1-(n-1)an=an.当n3时,nan+1=nan,即an+1=an.又a9=
6、8,a3=8.又2a3=a1+a2,a1=7,a2=2a3-a1=9.答案95.(2020陕西西安高三二模)数列an满足a1+2a2+3a3+nan=2n-1(nN+),则an=.若存在nN+使得ann+1n成立,则实数的最小值为.解析当n2时,a1+2a2+3a3+(n-1)an-1+nan=2n-1,a1+2a2+3a3+(n-1)an-1=2n-1-1,两式相减得nan=(2n-1)-(2n-1-1)=2n-1,所以an=2n-1n(n2).当n=1时,a1=1满足上式,综上所述,an=2n-1n.存在nN+使得ann+1n成立的充要条件为存在nN+使得2n-1n+1,设bn=2n-1n
7、+1,所以bn+1bn=2nn+22n-1n+1=2(n+1)n+21,即bn+1bn,所以bn单调递增,bn的最小项b1=12,即有b1=12,的最小值为12.答案an=2n-1n126.(2020广州高三月考)已知数列an满足:a1=2,an+an-1=4n-2(n2).(1)求数列an的通项公式;(2)若数列bn满足b1+3b2+7b3+(2n-1)bn=an,求数列bn的通项公式.解(1)由an+an-1=4n-2(n2)可化为(an-2n)+(an-1-2n+2)=0.令cn=an-2n,则cn+cn-1=0,即cn=-cn-1.因为a1=2,所以c1=a1-2=0,所以cn=0,即
8、an-2n=0,故an=2n.(2)由b1+3b2+7b3+(2n-1)bn=an,可知b1+3b2+7b3+(2n-1-1)bn-1=an-1(n2),两式作差得(2n-1)bn=an-an-1=2(n2),即bn=22n-1(n2).又当n=1时,b1=a1=2也满足上式,故bn=22n-1.素养培优练(2020江西师大附中高三一模)已知数列an中,a1=1,其前n项和为Sn,且满足2Sn=(n+1)an(nN+).(1)求数列an的通项公式;(2)记bn=3n-an2,若数列bn为递增数列,求的取值范围.解(1)2Sn=(n+1)an,2Sn+1=(n+2)an+1,2an+1=(n+2)an+1-(n+1)an,即nan+1=(n+1)an,an+1n+1=ann,ann=an-1n-1=a11=1,an=n(nN+).(2)由(1)知bn=3n-n2.bn+1-bn=3n+1-(n+1)2-(3n-n2)=23n-(2n+1).数列bn为递增数列,23n-(2n+1)0,即为递增数列,c1=2,即的取值范围为(-,2).
