專題六數(shù)列6.1數(shù)列的概念及表示五年高考考點(diǎn)數(shù)列的概念及表示1.(2022全國(guó)乙理,4,5分,中)嫦娥二號(hào)衛(wèi)星在完成探月任務(wù)后,繼續(xù)進(jìn)行深空探測(cè),成為我國(guó)第一顆環(huán)繞太陽(yáng)飛行的人造行星.為研究嫦娥二號(hào)繞日周期與地球繞日周期的比值,用到數(shù)列{bn}:b1=1+1α1,b2=1+1α1+1α2,b3=1+1α1+1α2+1α3,…,依此類推,其中αA.b1<b5B.b3<b8C.b6<b2D.b4<b7答案D2.(2021浙江,10,5分,難)已知數(shù)列{an}滿足a1=1,an+1=an1+an(n∈N*).記數(shù)列{an}的前n項(xiàng)和為Sn,則A.32<S100<3B.3<S100C.4<S100<92D.92答案A3.(2019上海,8,5分,易)已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿足Sn+an=2,則S5=.

答案314.(2014課標(biāo)Ⅱ,16,5分,易)數(shù)列{an}滿足an+1=11?an,a8=2,則a1答案15.(2015江蘇,11,5分,中)設(shè)數(shù)列{an}滿足a1=1,且an+1-an=n+1(n∈N*),則數(shù)列1an前10項(xiàng)的和為答案206.(2022北京,15,5分,難)已知數(shù)列{an}的各項(xiàng)均為正數(shù),其前n項(xiàng)和Sn滿足an·Sn=9(n=1,2,…).給出下列四個(gè)結(jié)論:①{an}的第2項(xiàng)小于3;②{an}為等比數(shù)列;③{an}為遞減數(shù)列;④{an}中存在小于1100的項(xiàng)其中所有正確結(jié)論的序號(hào)是.

答案①③④7.(2016課標(biāo)Ⅲ,17,12分,中)已知各項(xiàng)都為正數(shù)的數(shù)列{an}滿足a1=1,an2-(2an+1-1)an-2an+1=0.(1)求a2,a3;(2)求{an}的通項(xiàng)公式.解析(1)令n=1,則由題可得a12-(2a2-1)a1-2a2把a(bǔ)1=1代入,得1-(2a2-1)-2a2=0,解得a2=12令n=2,則a22-(2a3-1)a2-2a3即14?12(2a3-1)-2a3=0,解得a(2)將an2-(2an+1-1)an-2an+1=0因式分解可得(an+1)·(an-2an+1)=0,則an=-1(舍)或an-2an+1即an+1an=12,即{an}是以1為首項(xiàng),12三年模擬基礎(chǔ)強(qiáng)化練1.(2025屆安徽合肥一六八中學(xué)月考,3)已知數(shù)列{an}滿足an+1(1-an)=1,若a1=-1,則a10=()A.2B.-2C.-1D.1答案C2.(2025屆廣東汕頭摸底考,5)已知數(shù)列an=n?2025n?2024(n∈N*),則數(shù)列{an}A.a1,a100B.a45,a44C.a45,a1D.a44,a100答案B3.(2025屆遼寧沈陽(yáng)聯(lián)考,6)已知數(shù)列{an}的前n項(xiàng)和為Sn.若an+an+1=2n+5,a1=1,則S8=()A.48B.50C.52D.54答案C4.(2025屆重慶西南大學(xué)附中階段性檢測(cè),7)已知數(shù)列{an}的首項(xiàng)a1=2025,其前n項(xiàng)和Sn滿足Sn=n2an,則a2024=()A.1答案C5.(2025屆浙江名校聯(lián)考,6)數(shù)列{an}滿足an+2=2an+1+3an,則下列a1,a2的值能使數(shù)列{an}為周期數(shù)列的是()A.a1=0,a2=1B.a1=-1,a2=1C.a1=0,a2=2D.a1=-2,a2=0答案B6.(2025屆遼寧沈陽(yáng)東北育才學(xué)校開(kāi)學(xué)考,12)若數(shù)列{an}滿足a1=1,an+1-an=n+1,數(shù)列1an的前n項(xiàng)和為Sn,則S2024=答案40487.(2025屆安徽江淮十校聯(lián)考,13)記Sn為數(shù)列{an}的前n項(xiàng)和.已知3Snn+n=3an+1,a1=23,則數(shù)列{an答案an=238.(2024浙江寧波模擬,18)已知數(shù)列{an}滿足a1=1,且對(duì)任意正整數(shù)m,n都有am+n=an+am+2mn.(1)求數(shù)列{an}的通項(xiàng)公式;(2)求數(shù)列{(-1)nan}的前n項(xiàng)和Sn.解析(1)由對(duì)任意正整數(shù)m,n均有am+n=an+am+2mn,取m=1,得an+1=an+1+2n,當(dāng)n≥2時(shí),an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1+3+5+…+2n-1=n(1+2n?1)2當(dāng)n=1時(shí),a1=1,符合上式,所以an=n2,n∈N*.(2)當(dāng)n為偶數(shù)時(shí),Sn=(-12+22)+(-32+42)+…+[-(n-1)2+n2]=3+7+11+…+(2n-1)=n2當(dāng)n為奇數(shù)時(shí),Sn=Sn-1+(-1)nan=Sn-1-an=(n綜上所述:Sn=n

能力拔高練1.(2025屆福建寧德福鼎一中月考,8)已知正項(xiàng)數(shù)列{bn}滿足b1=1,bn=nbn+12nA.b2=1+52B.{bnC.bn+1-bn<1答案C2.(2024遼寧大連期末,8)已知函數(shù)f(x),若數(shù)列an=f(n),n∈N*為遞增數(shù)列,則稱函數(shù)f(x)為“數(shù)列保增函數(shù)”,已知函數(shù)f(x)=-ln(2x)+λx為“數(shù)列保增函數(shù)”,則λ的取值范圍是()A.ln32,+∞B.(C.[1,+∞)D.1答案B3.(2025屆廣東一調(diào),7)斐波那契數(shù)列因數(shù)學(xué)家斐波那契以兔子繁殖為例而引入,又稱“兔子數(shù)列”.這一數(shù)列如下定義:設(shè){an}為斐波那契數(shù)列,a1=1,a2=1,an=an-1+an-2(n≥3,n∈N*),其通項(xiàng)公式為an=151+52n?1?52n,設(shè)n是log2[(1+5)x-(1-5)x]<A.5B.6C.7D.8答案A4.(2025屆廣東深圳開(kāi)學(xué)考,8)數(shù)列{an}中,a1=2,an+1=an2-an+1,記An=1a1+1a2+…+1aA.A2024+B2024>1B.A2024+B2024<1C.A2024-B2024>1答案C5.(多選)(2025屆山東青島五十八中期中,10)若數(shù)列{an}滿足a1=1,a2=1,an=an-1+an-2(n≥3,n∈N*),則稱數(shù)列{an}為斐波那契數(shù)列,又稱黃金分割數(shù)列,則下列結(jié)論成立的是()A.a7=13B.2an=an-2+an+2(n≥3,n∈N*)C.a1+a3+a5+…+a2023=a2024D.a2+a4+a6+…+a2024=a2025答案AC6.(2025屆云南名校聯(lián)考,13)自然常數(shù)e是自然對(duì)數(shù)的底數(shù),大約等于2.71828.某人用“調(diào)日法”找逼近e的分?jǐn)?shù),稱小于2.718281的值為弱值,大于2.718282的值為強(qiáng)值.由21<e<31,取2為弱值,3為強(qiáng)值,得a1=2+31+1=52,故a1為弱值,與上一次的強(qiáng)值3計(jì)算得a2=5+32+1=83,故a2為弱值,繼續(xù)計(jì)算,……,若某次得到的近似值為弱值,與上一次的強(qiáng)值繼續(xù)計(jì)算得到新的近似值;若某次得到的近似值為強(qiáng)值,與上一次的弱值繼續(xù)計(jì)算得到新的近似值,答案67.(2025屆河北邯鄲第一次調(diào)研,14)已知有窮遞增數(shù)列{an}的各項(xiàng)均為正整數(shù)(n≥3),所有項(xiàng)的和為S,所有項(xiàng)的積為T,若T=4S,則該數(shù)列可能為.(填寫一個(gè)數(shù)列即可)

答案1,5,24(答案不唯一)8.(2025屆北京清華大學(xué)附屬中學(xué)開(kāi)學(xué)考,15)已知數(shù)列{an}滿足a