备战2025年高考二轮复习数学课件专题：数列 专题突破练-数列的递推关系.pptxVIP

数列的递推关系

1234567891011121314一、选择题1.(2024·陕西西安模拟)已知数列{an}满足,则a2024=()A.2024 B.2023 C.4047 D.4048C

12345678910111213142.(2024·湖北黄冈模拟)数列{an}的首项为1,前n项和为Sn,若Sn+Sm=Sn+m(m,n∈N*),则a9=()A.9 B.1 C.8 D.45B解析由题意知,数列{an}的首项为1,且Sn+Sm=Sn+m,令m=1,可得Sn+S1=Sn+1,即Sn+1-Sn=S1=1,所以数列{Sn}是首项为1,公差为1的等差数列,所以Sn=1+(n-1)×1=n,则a9=S9-S8=1.故选B.

1234567891011121314A解析因为3Sn=an+1,则3Sn+1=an+1+1,两式相减可得3an+1=an+1-an,即2an+1=-an,令n=7,可得2a8=-a7,且an≠0,所以.故选A.

12345678910111213144.(2024·湖北武汉模拟)设等比数列{an}的前n项和为Sn,已知Sn+1=3Sn+2,n∈N*,则S5=()A.80 B.160 C.121 D.242D

12345678910111213145.(2024·河北唐山二模)已知数列{an}满足an+1=an+a1+2n,a10=130,则a1=()A.1 B.2 C.3 D.4D解析由题意可得an+1-an=a1+2n,则可得a2-a1=a1+2,a3-a2=a1+4,…,a10-a9=a1+18,将以上等式左右两边分别相加得,a10-a1=9a1+=9a1+90,即a10=10a1+90.又a10=130,所以a1=4.故选D.

12345678910111213146.(2024·山东青岛模拟)若数列{an}满足(n-1)an=(n+1)an-1(n≥2),a1=2,则满足不等式an930的最大正整数n为()A.28 B.29 C.30 D.31B所以an=n(n+1),{an}是递增数列.由an=n(n+1)930,(n+31)(n-30)0,解得-31n30,所以n的最大值为29.故选B.

1234567891011121314A

1234567891011121314

12345678910111213148.(2024·安徽合肥模拟)数学家斐波那契在研究兔子繁殖问题时,发现有这样一个数列{an}:1,1,2,3,5,8,…,其中从第3项起,每一项都等于它前面两项之和,即a1=a2=1,an+2=an+1+an,这样的数列称为“斐波那契数列”.若am=2(a3+a6+a9+…+a174)+1,则m=()A.175 B.176 C.177 D.178B

解析由从第三项起,每个数等于它前面两个数的和,a1=a2=1,由an+2=an+1+an(n∈N*),得an=an+2-an+1,所以a1=a3-a2,a2=a4-a3,a3=a5-a4,…,an=an+2-an+1,将这n个式子左右两边分别相加可得Sn=a1+a2+…+an=an+2-a2=an+2-1,所以Sn+1=an+2.所以2(a3+a6+a9+…+a174)+1=(a3+a3+a6+a6+a9+a9+…+a174+a174)+1=a1+a2+a3+a4+a5+a6+a7+a8+a9+…+a172+a173+a174+1=S174+1=a176.故m=176.故选B.1234567891011121314

1234567891011121314二、选择题AB

1234567891011121314

1234567891011121314BC

1234567891011121314

1234567891011121314

1234567891011121314三、填空题11.(2024·四川泸州三模)已知Sn是数列{an}的前n项和,a1=1,nan+1=(n+2)Sn,则an=.?(n+1)·2n-2

1234567891011121314

123456789101112131412.设数列{an}的前n项和为Sn,且Sn=2an+n-7,若30ak50,则k的值为.?4

1234567891011121314解析因为Sn=2an+n-7,①所以当n=1时,S1=2a1+1-7=a1,解得a1=6.又Sn-1=2an-1+n-1-7(n≥2),②①-②得an=2an-2an-1+1,即an=2an-1-1(n≥2),所以an-1=2(an-1