備戰(zhàn)2025年高考二輪復(fù)習(xí)數(shù)學(xué)專題突破練11

專題突破練(分值:92分)學(xué)生用書P163一、選擇題:本題共8小題,每小題5分,共40分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的.1.(2024·陜西西安模擬)已知數(shù)列{an}滿足∑k=1nak2k-1=n+A.2024 B.2023 C.4047 D.4048答案C解析由題意可得a1+a23+a35+…+an2n-1=n+1,當(dāng)n=1時(shí),a1=2;當(dāng)n≥2時(shí),a1+a23+a35+綜上所述,an=2,n所以a2024=4047,故選C.2.(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.45答案B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.3.(2024·福建漳州一模)已知各項(xiàng)均不為0的數(shù)列{an}的前n項(xiàng)和為Sn,若3Sn=an+1,則a8a7=A.-12 B.-13 C.答案A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a8a7=-14.(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.242答案D解析由Sn+1=3Sn+2,得Sn=3Sn-1+2(n≥2),所以Sn+1-Sn=3Sn-3Sn-1,得an+1=3an,所以等比數(shù)列{an}的公比為q=3,所以由Sn+1=3Sn+2,得a1(1-3n+1)1-3=3×a1(1-3n)1-3+2,所以所以S5=2×(1-35)1-3=35.(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.4答案D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+9×(2+18)2=9a1+90,即a10=10a又a10=130,所以a1=4.故選D.6.(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.31答案B解析依題意,數(shù)列{an}滿足(n-1)an=(n+1)an-1(n≥2),a1=2,anan-1=n+1n-1(n≥2),所以an=a1·a2a1·a3a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.7.(2024·陜西咸陽三模)某校組織知識(shí)競(jìng)賽,已知甲同學(xué)答對(duì)第一題的概率為911,從第二題開始,甲同學(xué)回答第n題時(shí)答錯(cuò)的概率為Pn,Pn=112Pn-1+23,當(dāng)n≥2時(shí),Pn≥M恒成立,則M的最大值為A.1522 B.1722答案A解析依題意,P1=1-911=211,當(dāng)n≥2時(shí),由Pn=112Pn-1+23,得Pn-811=112Pn-1-811,而P1-811=-611,因此數(shù)列Pn-811是首項(xiàng)為-611,公比為112的等比數(shù)列,則Pn-811=-611×112n-1,即Pn=811-611×112n-1,8.(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.178答案B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.二、選擇題:本題共2小題,每小題6分,共12分.在每小題給出的選項(xiàng)中,有多項(xiàng)符合題目要求,全部選對(duì)的得6分,部分選對(duì)的得部分分,有選錯(cuò)的得0分.9.(2024·河北石家莊模擬)已知數(shù)列{an}滿足a1=13,an+1=2an2an+1(n∈A.1B.{an}的通項(xiàng)公式為an=2C.{an}為遞減數(shù)列D.1an的前n項(xiàng)和Tn=2n+1答案AB解析因?yàn)閍n+1=2an2an+1(n∈N*),由題意顯然an≠0,an+1≠0,變形得1an+1=1+2an2an=12×1an+1,所以1an+1-2=12(1an-2).又因?yàn)?a1-2=1≠0,所以1an-2是以1為首項(xiàng),12為公比的等比數(shù)列,A正確;因?yàn)?an-2=(12)n-1,所以an=12+(12)

n-1=2n-12n+1,B正確;因?yàn)?12)n-1單調(diào)遞減,所以an=12+(12)10.(2024·山東濟(jì)寧三模)已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿足2Sn=3n+1-3,數(shù)列{bn}的前n項(xiàng)和為Tn,且滿足Tnn=12bn+1,A.a1=3b1B.數(shù)列{an}是等比數(shù)列C.數(shù)列{bn}是等差數(shù)列D.若b2=3,則∑答案BC解析由2Sn=3n+1-3,得Sn=12×3n+1-32,a1=S1=3.當(dāng)n≥2時(shí),an=Sn-Sn-1=12(3n+1-3n)=3n,a1=3滿足上式,因此an=3n,數(shù)列{an}是等比數(shù)列,B正確;由Tnn=12bn+1,得Tn=n2bn+n,b1=T1=12b1+1,解得b1=2,a1≠3b1,A錯(cuò)誤;當(dāng)n≥2時(shí),Tn-1=n-12bn-1+n-1,兩式相減得n-22bn+1=n-12bn-1,于是n2bn=n-12bn+1+1,兩式相加得2n-22bn=n-12bn-1+n-12bn+1,整理得2bn=bn-1+bn+1,因此數(shù)列{bn}是等差數(shù)列,C正確;當(dāng)b2=3時(shí),三、填空題:本題共2小題,每小題5分,共10分.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=.

答案(n+1)·2n-2解析當(dāng)n≥2時(shí),(n-1)an=(n+1)Sn-1,即Sn=nn+2an+1,Sn-1=n-則Sn-Sn-1=nn+2an+1-n-1n即an+1an=2(n+2)n+1,則有anan-1=2(n當(dāng)n=1時(shí),a1=1,符合上式,故an=(n+1)·2n-2.12.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.

答案4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an-1an-1-1=2,所以{a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,所以295<2k-1<495,又k∈N*,所以k=四、解答題:本題共2小題,共30分.解答應(yīng)寫出文字說明、證明過程或演算步驟.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.解(1)4Sn=(an+1)2,當(dāng)n≥2時(shí),4Sn-1=(an-1+1)2,兩式作差可得4an=an2-an-12+2an-2an-1,(an-an-1)(an+an-1)-2(an+an-1)=0,(an+an-1)(an又an>0,故an-an-1=2.又當(dāng)n=1時(shí),4a1=(a1+1)2,解得a1=1.故數(shù)列{an}為首項(xiàng)為1,公差為2的等差數(shù)列,則an=2n-1.(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)=2n+23[1-4n]14.(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an+1=-12an2+32,證明:-1<an<1(1)解因?yàn)?Sn=2-2an+1且a1=1,當(dāng)n=1時(shí),3S1=2-2a2,解得a2=-1當(dāng)n≥2時(shí),3Sn-1=2-2an,所以3Sn-3Sn-1=2-2an+1-(2-2an),即3an=2an-2an+1,所以an+1an=-1又a2a1=-12,所以{an}是以1為首項(xiàng),-12為公比的等比數(shù)列,所以an=-12n-因?yàn)?an+2-an+1)(an+2-an)=[-12n+1--12n][-12n+1--12n-1]=[-12×-12n--1所以{an}是“跳動(dòng)數(shù)列”.(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-anq)(anq2-an)<0,即an2(q2-q)(q2-1)<0,所以(q2-q)(q2-1)<0,即q(q-1)2(q+1)<0,解得-1<q<0,即實(shí)數(shù)q的取值范圍為(-(3)證明由an+1=-12可得an+2=-12所以an+2-an+1=-12an+12+32--12則an+2-an=(an+2-an+1)+(an+1-an)=12(an-an+1)(an+an+1)+(an+1-an)=(an+1-an)-12(an+an+1)+1=-12(a由{