数列的概念与简单表示法+课时突破练课件-2027届高考数学一轮复习

数学课时突破练37数列的概念与简单表示法12345678910111213141516基础•满分练1.(2025·江西赣州期中)已知数列{an}的前4项为3,6,11,20,则{an}的通项公式可以是(

)A.an=3n B.an=2n+nC.an=3n

D.an=4n-nB解析:已知数列{an}的前4项为3,6,11,20,即a1=3,a2=6,a3=11,a4=20,验证选项中的通项公式,只有an=2n+n符合题意,所以{an}的通项公式可以是an=2n+n.故选B.123456789101112131415162.(2025·北京大兴期中)已知数列{an}满足a1=1,an=an-1+n(n≥2),则a4=(

)A.5 B.10

C.11

D.12B解析:因为数列{an}满足a1=1,an=an-1+n(n≥2),则a2=a1+2=3,a3=a2+3=6,a4=a3+4=10.故选B.123456789101112131415163.(2025·广东期末)已知数列{an}的前n项和Sn=kn2+2n,a2=5,则k的值为(

)A.2 B.-2 C.1

D.-1C解析:由Sn=kn2+2n,可得a2=S2-S1=(4k+4)-(k+2)=3k+2,则3k+2=5,解得k=1.故选C.123456789101112131415164.(2025·四川凉山期中)已知数列{an}满足a1=4,且an+1=2an-3,则a211=(

)A.2210-3 B.2211+3C.2210+3 D.2211+1C解析:因为an+1=2an-3,所以an+1-3=2(an-3).因为a1-3=1,所以数列{an-3}是首项为1,公比为2的等比数列,所以an-3=2n-1,所以an=2n-1+3,故a211=2210+3.故选C.12345678910111213141516

A

12345678910111213141516

A12345678910111213141516

12345678910111213141516

AC

123456789101112131415168.(原创)已知数列{an}中,a1=4,(n+1)an+1=(n+2)an,则an=

.2n+2

12345678910111213141516

C

1234567891011121314151610.(2025·湖南常德一模)已知数列{an}的前n项和为Sn,a1=1,且an+1=a1+a2+…+an(n∈N*),则(

)A.a2=2 B.a4=8C.S2=3 D.S5=16能力•高分练D12345678910111213141516

1234567891011121314151611.(2025·安徽滁州一模)已知数列{an}的第1项和第2项均为1,以后各项由an+2=an+1+an(n∈N*)给出.若数列{an}的各项除以3所得余数组成一个新数列{bn},则b2024+b2025=(

)A.1 B.2

C.3

D.4A解析:因为an+2=an+an+1(n∈N*),a1=a2=1,所以数列{an}为1,1,2,3,5,8,13,21,34,55,89,144,…,此数列各项除以3的余数依次构成的数列{bn}为1,1,2,0,2,2,1,0,1,1,2,0,2,2,1,0,…,所以{bn}是以8为周期的周期数列,所以b2

024+b2

025=0+1=1.故选A.12345678910111213141516

ACD12345678910111213141516

1234567891011121314151613.(2026·江苏南京模拟)已知数列{an}的前n项和Sn=3n2-2n+1,则an=

.

12345678910111213141516

12345678910111213141516素养•提升练

A12345678910111213141516

123456789101112