2025高考數(shù)學(xué)二輪復(fù)習(xí)-專題突破練11 數(shù)列的遞推關(guān)系【課件】

專題突破練11數(shù)列的遞推關(guān)系20251234567891011121314一、選擇題1.(2024·陜西西安模擬)已知數(shù)列{an}滿足,則a2024=(

)A.2024 B.2023 C.4047 D.4048C12345678910111213142.(2024·湖北黃岡模擬)數(shù)列{an}的首項(xiàng)為1,前n項(xiàng)和為Sn,若Sn+Sm=Sn+m(m,n∈N*),則a9=(

)A.9 B.1 C.8 D.45B解析

由題意知,數(shù)列{an}的首項(xiàng)為1,且Sn+Sm=Sn+m,令m=1,可得Sn+S1=Sn+1,即Sn+1-Sn=S1=1,所以數(shù)列{Sn}是首項(xiàng)為1,公差為1的等差數(shù)列,所以Sn=1+(n-1)×1=n,則a9=S9-S8=1.故選B.1234567891011121314A解析

因?yàn)?Sn=an+1,則3Sn+1=an+1+1,兩式相減可得3an+1=an+1-an,即2an+1=-an,令n=7,可得2a8=-a7,且an≠0,所以.故選A.12345678910111213144.(2024·湖北武漢模擬)設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,已知Sn+1=3Sn+2,n∈N*,則S5=(

)A.80 B.160

C.121

D.242D12345678910111213145.(2024·河北唐山二模)已知數(shù)列{an}滿足an+1=an+a1+2n,a10=130,則a1=(

)A.1 B.2 C.3 D.4D解析

由題意可得an+1-an=a1+2n,則可得a2-a1=a1+2,a3-a2=a1+4,…,a10-a9=a1+18,將以上等式左右兩邊分別相加得,a10-a1=9a1+=9a1+90,即a10=10a1+90.又a10=130,所以a1=4.故選D.12345678910111213146.(2024·山東青島模擬)若數(shù)列{an}滿足(n-1)an=(n+1)an-1(n≥2),a1=2,則滿足不等式an<930的最大正整數(shù)n為(

)A.28 B.29 C.30 D.31B所以an=n(n+1),{an}是遞增數(shù)列.由an=n(n+1)<930,(n+31)(n-30)<0,解得-31<n<30,所以n的最大值為29.故選B.1234567891011121314A123456789101112131412345678910111213148.(2024·安徽合肥模擬)數(shù)學(xué)家斐波那契在研究兔子繁殖問題時(shí),發(fā)現(xiàn)有這樣一個(gè)數(shù)列{an}:1,1,2,3,5,8,…,其中從第3項(xiàng)起,每一項(xiàng)都等于它前面兩項(xiàng)之和,即a1=a2=1,an+2=an+1+an,這樣的數(shù)列稱為“斐波那契數(shù)列”.若am=2(a3+a6+a9+…+a174)+1,則m=(

)A.175 B.176

C.177 D.178B解析

由從第三項(xiàng)起,每個(gè)數(shù)等于它前面兩個(gè)數(shù)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,將這n個(gè)式子左右兩邊分別相加可得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.故選B.12345678910111213141234567891011121314二、選擇題AB12345678910111213141234567891011121314BC123456789101112131412345678910111213141234567891011121314三、填空題11.(2024·四川瀘州三模)已知Sn是數(shù)列{an}的前n項(xiàng)和,a1=1,nan+1=(n+2)Sn,則an=

.

(n+1)·2n-21234567891011121314123456789101112131412.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2an+n-7,若30<ak<50,則k的值為

.

41234567891011121314解析

因?yàn)镾n=2an+n-7,①所以當(dāng)n=1時(shí),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-1-1),又a1-1=5≠0,即

=2,所以{an-1}是以5為首項(xiàng),2為公比的等比數(shù)列,所以an-1=5·2n-1,即an=5·2n-1+1.因?yàn)?0<ak<50,所以30<5·2k-1+1<50,1234567891011121314四、解答題13.(15分)(2024·黑龍江哈爾濱模擬)已知各項(xiàng)均為正數(shù)的數(shù)列{an}滿足4Sn=(an+1)2,其中Sn是數(shù)列{an}的前n項(xiàng)和.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列{bn}滿足bn=(-1)nan+[(-1)n+1]2n,求{bn}的前2n項(xiàng)和T2n.1234567891011121314(2)bn=(-1)nan+[(-1)n+1]2n=(-1)n·(2n-1)+[(-1)n+1]×2n.b2k+b2k-1=(-1)2k·(4k-1)+[(-1)2k+1]×22k+(-1)2k-1(4k-3)+[(-1)2k-1+1]×22k-1=4k-1+22k+1-4k+3=2+22k+1,k∈N*,故T2n=b1+b2+b3+b4+…+b2n-1+b2n=(2+23)+(2+25)+…+(2+22n+1)123456789101112131414.(15分)(2024·山東菏澤模擬)定義:如果數(shù)列{an}從第三項(xiàng)開始,每一項(xiàng)都介于前兩項(xiàng)之間,那么稱數(shù)列{an}為“跳動(dòng)數(shù)列”.(1)若數(shù)列{an}的前n項(xiàng)和Sn滿足3Sn=2-2an+1,且a1=1,求{an}的通項(xiàng)公式,并判斷{an}是否為“跳動(dòng)數(shù)列”;(2)若公比為q的等比數(shù)列{an}是“跳動(dòng)數(shù)列”,求實(shí)數(shù)q的取值范圍;(3)若“跳動(dòng)數(shù)列”{an}滿足12345678910111213141234567891011121314(2)解

由“跳動(dòng)數(shù)列”的定義可知:{an}是“跳動(dòng)數(shù)列”?(an+2-an+1)(an+2-an)<0.若公比為q的等比數(shù)列{an}是“跳動(dòng)數(shù)列”,則(an+2-an+1)(an+2-an)=(anq2