2024-2025學(xué)年新教材高中數(shù)學(xué)第五章數(shù)列5.1數(shù)列基礎(chǔ)5.1.2數(shù)列中的遞推課時(shí)素養(yǎng)檢測(cè)含解析新人教B版選擇性必修第三冊(cè)

PAGE二數(shù)列中的遞推(30分鐘60分)一、選擇題(每小題5分,共30分,多選題全部選對(duì)得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)1.已知數(shù)列{an}的首項(xiàng)為a1=1,且an+1=QUOTEan+QUOTE,則此數(shù)列第4項(xiàng)是 ()A.1 B.QUOTE C.QUOTE D.QUOTE【解析】選A.a1=1,a2=QUOTE×1+QUOTE=1,a3=QUOTE×1+QUOTE=1,a4=QUOTE×1+QUOTE=1.2.符合遞推關(guān)系式an=QUOTEan-1(n≥2)的數(shù)列是 ()A.1,2,3,4,… B.1,QUOTE,2,2QUOTE,…C.QUOTE,2,QUOTE,2,… D.0,QUOTE,2,2QUOTE,…【解析】選B.由題意知從第2項(xiàng)起先每一項(xiàng)是前一項(xiàng)的QUOTE倍,只有B項(xiàng)符合.3.數(shù)列{an}中,a1=1,對(duì)全部的n≥2,n∈N*,都有a1·a2·a3·…·an=n2,則a3+a5等于 ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選C.a1a2a3=32,a1aa1a2a3a4a5=5則a3=QUOTE=QUOTE,a5=QUOTE=QUOTE.故a3+a5=QUOTE.4.已知數(shù)列{an}滿意:a1=QUOTE,對(duì)于隨意的n∈N*,an+1=QUOTEan(1-an),則a2019-a2020= ()A.-QUOTE B.QUOTE C.-QUOTE D.QUOTE【解析】選D.a1=QUOTE,a2=QUOTE×QUOTE×QUOTE=QUOTE,a3=QUOTE×QUOTE×QUOTE=QUOTE,a4=QUOTE×QUOTE×QUOTE=QUOTE,…,歸納可知,當(dāng)n>1時(shí),若n為奇數(shù),an=QUOTE;若n為偶數(shù),an=QUOTE,所以a2019-a2020=QUOTE-QUOTE=QUOTE.5.數(shù)列{an}的前n項(xiàng)和Sn=3n-2n2(n∈N*),則當(dāng)n≥2時(shí),下列不等式成立的是 ()A.Sn>na1>nan B.Sn>nan>na1C.na1>Sn>nan D.nan>Sn>na1【解析】選C.由an=QUOTE解得an=5-4n,a1=1,所以na1=n,所以nan=5n-4n2,因?yàn)閚a1-Sn=n-(3n-2n2)=2n2-2n=2n(n-1)>0(當(dāng)n≥2時(shí)).Sn-nan=3n-2n2-(5n-4n2)=2n2-2n>0(當(dāng)n≥2時(shí)).所以na1>Sn>nan.【補(bǔ)償訓(xùn)練】若數(shù)列{an}的前n項(xiàng)和是Sn=n2-4n+2,則|a1|+|a2|+…+|a10|等于 ()A.15B.35C.66【解析】選C.易得an=QUOTE|a1|=1,|a2|=1,|a3|=1,令an>0,則2n-5>0,所以n≥3.所以|a1|+|a2|+…+|a10|=1+1+a3+…+a10=2+(S10-S2)=2+[(102-4×10+2)-(22-4×2+2)]=66.6.(多選題)已知數(shù)列{an}滿意:an≤an+1,an=n2+λn,n∈N*,則實(shí)數(shù)λ可以取的值為 ()A.-3 B.-2 C.1 【解析】選ABCD.an≤an+1?n2+λn≤(n+1)2+λ(n+1)?λ≥-(2n+1),n∈N*?λ≥-3.二、填空題(每小題5分,共10分)7.設(shè)數(shù)列{an}滿意:a1=2,an+1=1-QUOTE,記數(shù)列{an}的前n項(xiàng)之積為Πn,則Π2021的值為________.

【解析】由a2=QUOTE,a3=-1,a4=2可知,數(shù)列{an}是周期為3的周期數(shù)列,從而Π2021=(-1)673×2×QUOTE=-1.答案:-18.數(shù)列QUOTE滿意a1=1,an-an-1=QUOTE(n≥2且n∈N*),則數(shù)列QUOTE的通項(xiàng)公式為an=__________.

【解析】an-an-1=QUOTE=QUOTE-QUOTE,an=(an-an-1)+(an-1-an-2)+(an-2-an-3)+…+(a2-a1)+a1,an=QUOTE-QUOTE+QUOTE-QUOTE+…+QUOTE-QUOTE+1=2-QUOTE,當(dāng)n=1時(shí)滿意.答案:2-QUOTE三、解答題(每小題10分,共20分)9.在數(shù)列{an}中,a1=1,an+1=QUOTEan.(1)寫出數(shù)列的前5項(xiàng);(2)猜想數(shù)列的一個(gè)通項(xiàng)公式.【解析】(1)a1=1,a2=QUOTEa1=QUOTE,a3=QUOTEa2=QUOTE×QUOTE=QUOTE,a4=QUOTEa3=QUOTE×QUOTE=QUOTE,a5=QUOTEa4=QUOTE×QUOTE=QUOTE.(2)由(1)知an=QUOTE(n∈N*).10.已知數(shù)列{an}滿意:a1=m(m為正整數(shù)),an+1=QUOTE若a6=1,求m全部可能的取值.【解析】若a5為奇數(shù),則3a5+1=1,a5=0(舍去).若a5為偶數(shù),則QUOTE=1,a5=2.若a4為奇數(shù),則3a4+1=2,a4=QUOTE(舍去).若a4為偶數(shù),則QUOTE=2,a4=4.若a3為奇數(shù),則3a3+1=4,a3=1,則a2=2,a1=4.若a3為偶數(shù),則QUOTE