蘇教版高中數(shù)學(xué)選擇性必修第一冊第4章數(shù)列專題強化練8含答案

專題強化練8數(shù)列通項公式的求法1.在數(shù)列{an}中,a1=0,an+1=an+ln1+1n,則{aA.an=lnnB.an=(n-1)ln(n+1)C.an=nlnnD.an=lnn+n-22.(2024江蘇高郵調(diào)研)在數(shù)列{an}中,a1=1,an+1=2anan+2(n∈A.13.(2023重慶巴蜀中學(xué)月考)在數(shù)列{an}中,a1=2,an2+an+1=3an+1(n∈N*),Sn是數(shù)列1aA.1-1C.1-14.(2024江蘇鎮(zhèn)江丹陽期中)已知數(shù)列{an}的前n項和Sn=n2+2n,cn=12,n=1,1an-1an,A.6B.7C.8D.95.(2024江蘇蘇州工業(yè)園星海實驗中學(xué)月考)設(shè)數(shù)列{an}的前n項和為Sn,a1=12,an+1=A.2019B.2020C.2021D.20226.(多選題)(2023江蘇南通如皋期末)已知Sn是數(shù)列{an}的前n項和,a1=1,an+1=[2+(-1)n]·an+1+(-1)n+1,則下列結(jié)論正確的是()A.a2n+1=3a2nB.數(shù)列{a2n-1+3}為等比數(shù)列C.a2n=4×3n-1-2D.S2n=4×3n-4n-47.(2024四川成都田家炳中學(xué)月考)已知數(shù)列{an}的首項為2,等比數(shù)列{bn}滿足bn=an+1an且b1012=1,則a8.(2024山東德州期中)設(shè)數(shù)列{an}滿足a1=-2,an+1=an+n·2n,則log2a1026=.

9.(2024江蘇蘇州常熟期中)設(shè)數(shù)列{an}的前n項和為Sn,且Sn=2an-2n+1,數(shù)列{bn}滿足bn=log2ann+1,若不等式1+1b2·1+1b410.(2023江蘇常州第一中學(xué)調(diào)研)已知正項數(shù)列{an}滿足an-an-1=an+an-1(n∈N*,n≥2),a1=1.數(shù)列{bn}滿足各項均不為0,b1=4,其前n項積為Tn=2(1)求數(shù)列{an}的通項公式;(2)設(shè)cn=log2bn,求數(shù)列{cn}的通項公式;(3)在(2)的條件下,記數(shù)列{(-1)nan}的前2m項和為S2m,求使得不等式S2m≥c1+c2+…+c10成立的正整數(shù)m的最小值.

答案與分層梯度式解析專題強化練8數(shù)列通項公式的求法1.A由已知得an+1-an=lnn+1所以an-an-1=lnn-ln(n-1),an-1-an-2=ln(n-1)-ln(n-2),……a3-a2=ln3-ln2,a2-a1=ln2-ln1,將以上(n-1)個式子相加,整理得an-a1=lnn-ln1=lnn,又因為a1=0,所以an=lnn.故選A.2.D∵an+1=2anan+2,∴an+1(an+2)=2an,即an+1an兩邊同時除以an+1an,得1+2an=令bn=2an,則bn+1-bn=1,則{bn}是首項為b1=則bn=2+(n-1)=n+1,即2an=n+1,則an=2n+1,則a10=23.B因為an2+an+1=3an+1,所以an2+an-2=3an+1-3,即(an+2)(a兩邊同時取倒數(shù)得1(整理得1an-1所以S2023=1a1+2+1a2+24.B當(dāng)n=1時,a1=S1=3;當(dāng)n≥2時,an=Sn-Sn-1=(n2+2n)-[(n-1)2+2(n-1)]=2n+1,又a1=3符合上式,所以an=2n+1.當(dāng)n=1時,c1=12當(dāng)n≥2時,cn=1a所以c1+c2+c3+…+cm=12+1213-15.C由an+1=2anan+1,取倒數(shù)得1an+1=12故an=1-12n-1+1,故S2023=2023-12令M=120+1由12n<12n-1由12n-1+1<所以2021+122022<S2023<2021+故S2023∈(2021,2022),則正整數(shù)k的值為2021.故選C.方法總結(jié)若數(shù)列相鄰兩項an+1與an滿足an+1=pan+q,則可考慮設(shè)an+1+x=p(an+x)(其中x為待定系數(shù)),構(gòu)造以a1+x為首項,p為公比的等比數(shù)列{an+x},進(jìn)一步求{an}的通項公式.6.ABD當(dāng)n為奇數(shù)時,an+1=(2-1)an+1+1,則an+1=an+2,當(dāng)n為偶數(shù)時,an+1=3an+1-1=3an,對于A,由2n為偶數(shù),得a2n+1=3a2n,故A正確;對于B,當(dāng)n為偶數(shù)時,an+1=3an,用2n替換上式中的n得a2n+1=3a2n,當(dāng)n為奇數(shù)時,an+1=an+2,用2n-1替換上式中的n得a2n=a2n-1+2,則a2n+1=3(a2n-1+2),兩邊同時加3得a2n+1+3=3(a2n-1+3),又a1+3=4,所以數(shù)列{a2n-1+3}是首項為4,公比為3的等比數(shù)列,故B正確;對于C,當(dāng)n為奇數(shù)時,an+1=an+2,用2n+1替換上式中的n得a2n+2=a2n+1+2,又a2n+1=3a2n,所以a2n+2=3a2n+2,兩邊同時加1得a2n+2+1=3(a2n+1),因為a2=a1+2=3,所以a2+1=4,所以{a2n+1}是以4為首項,3為公比的等比數(shù)列,則a2n+1=4×3n-1,則a2n=4×3n-1-1,n∈N*,故C錯誤;對于D,由B得a2n-1+3=4×3n-1,則a2n-1=4×3n-1-3,所以S2n=(a1+a3+…+a2n-1)+(a2+a4+…+a2n)=(4+12+…+4×3n-1-3n)+(4+12+…+4×3n-1-n)=4(1-3n)1-3×2-4n=4×3n故選ABD.7.答案2解析設(shè)等比數(shù)列{bn}的公比為q,則bn=b1·qn-1=an所以anan-1=b1·qn-2,an-1an-2=b1·q累乘得anan-1·an-1an-2·所以a2024=a1·b12023·q2022×20232=2b12023·又b1012=b1·q1011=1,所以a2024=2(b1·q1011)2023=2.8.答案1036解析∵an+1=an+n·2n,∴an+1-an=n·2n,當(dāng)n≥2時,an-a1=an-an-1+an-1-an-2+…+a2-a1=(n-1)·2n-1+(n-2)·2n-2+…+2·22+1·21,①∴2(an-a1)=(n-1)·2n+(n-2)·2n-1+…+2·23+1·22,②①-②得-(an-a1)=-(n-1)·2n+2n-1+2n-2+…+23+22+2=-(n-1)·2n+2(1-2n-1)1-2=-(n-1)·2n-2+2n∴an-a1=(n-2)·2n+2,當(dāng)n=1時也符合上式,所以an=(n-2)·2n,故log2a1026=log2(1024×21026)=log2(210×21026)=log221036=1036.9.答案3解析因為Sn=2an-2n+1,所以Sn-1=2an-1-2n(n≥2),兩式相減得an=2an-2an-1-2n,即an=2an-1+2n,兩邊同除以2n得an又當(dāng)n=1時,a1=2a1-22,所以a1=4,所以a1所以an所以an2n=2+(n-1)×1=n+1,故an=(n+1)·所以bn=log2ann+1=log2代入題中不等式得1+121+14…1+令cn=1+1則cn+1=1+1所以cn+1-cn=1+=1+=1+1因為(2n+3)2-(2n+2)(2n+4)=1>0,所以(2n+3)2>(2n+2)(2n+4),所以2n+3>2n所以cn+1-cn>0恒成立,即{cn}為遞增數(shù)列,所以(cn)min=c1=1+1所以m≤34,即m的最大值為310.解析(1)∵an-an-1=an+an-1(n≥2),∴{an}是首項為a1=1,公差為1的等差數(shù)列,則an=1+(n-1)×1=n,故an(2)當(dāng)n=1時,T1=b2,即b2=b1=4,當(dāng)n≥2時,Tn-1=2n-2·bn,則bn=TnTn-1=則log2bn+1=log212bn2=2log2bn-1,即cn+1=2cn-1,∴cn+1-1=2(c又∵c1-1=log2b1-1=1,c2-1=log2b2-1=1,∴當(dāng)n=1時不滿足上式,故當(dāng)n≥2時,{cn-1}是首項為1,公比為2的等比數(shù)列,∴cn-1=1×2n-2,則cn=1+2n-2(n≥2),綜上所述,cn=2,(3)∵(-1)2k-1a2k-1+(-1)2ka2k=-(2k-1)2+(2k)2=(2k+2k-1)[2k-(2k-1)]=4k-1,∴S2m=4(1+2+…+m)-m=4×(1+m)m又∵c1+c2+…+c10=2+9+(1+2+…+28)=11+1×(1-29)1-2=522,∴2m∵y=2x2+x在[1,+∞)上單調(diào)遞增,2×162+16=528>522,2×152+15=465<522,且m∈N*,