2022高考数学一轮复习高考大题专项练三数列文含解析北师大版.docx

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1、高考大题专项(三)　数列1.(2020某某天河区模拟)已知Sn为数列{an}的前n项和,且a1<2,an>0,6Sn=an2+3an+2,n∈N+.(1)求数列{an}的通项公式;(2)若任意n∈N+,bn=(-1)nan2,求数列{bn}的前2n项和T2n.2.(2020某某某某二模,文17)设等差数列{an}的公差为d(d>1),前n项和为Sn,等比数列{bn}的公比为q.已知b1=a1,b2=3,2q=3d,S10=100.(1)求数列{an},{bn}的通项公式;(2)记=an·bn,求数列{}的前n项和Tn.3.(2020某某某某4月质检二,理17)

2、已知等差数列{an}的前n项和为Sn,a2=1,S7=14,数列{bn}满足b1·b2·b3·…·bn=2n2+n2.(1)求数列{an}和{bn}的通项公式;(2)若数列{}满足=bncos(anπ),求数列{}的前2n项和T2n.4.(2020某某某某二模)Sn为等比数列{an}的前n项和,已知a4=9a2,S3=13,且公比q>0.(1)求an及Sn.(2)是否存在常数λ,使得数列{Sn+λ}是等比数列?若存在,求出λ的值;若不存在,请说明理由.5.(2020某某,19)已知{an}为等差数列,{bn}为等比数列,a1=b1=1,a5=5(a4-a3),

3、b5=4(b4-b3).(1)求{an}和{bn}的通项公式;(2)记{an}的前n项和为Sn,求证:SnSn+20,所以an-an-1=3,所以数列{

4、an}是首项为1,公差为3的等差数列,所以an=1+3(n-1)=3n-2.(2)bn=(-1)nan2=(-1)n(3n-2)2.所以b2n-1+b2n=-(6n-5)2+(6n-2)2=36n-21.所以数列{bn}的前2n项的和T2n=36×(1+2+…+n)-21n=36×n(n+1)2-21n=18n2-3n.2.解(1)由题意,得S10=10a1+45d=100,b2=b1q=3,将b1=a1,q=32d代入上式,可得2a1+9d=20,a1d=2,解得a1=9,d=29(舍去),或a1=1,d=2.∴数列{an}的通项公式为an=1+2(n-1)

5、=2n-1,n∈N+.∴b1=a1=1,q=32d=32×2=3,∴数列{bn}的通项公式为bn=1×3n-1=3n-1,n∈N+.(2)由(1)知,=an·bn=(2n-1)·3n-1,∴Tn=c1+c2+c3+…+=1×1+3×3+5×32+…+(2n-1)·3n-1,①3Tn=1×3+3×32+…+(2n-3)·3n-1+(2n-1)·3n,②①-②,得-2Tn=1+2×3+2×32+…+2·3n-1-(2n-1)·3n=1+2×(3+32+…+3n-1)-(2n-1)·3n=1+2×3-3n1-3-(2n-1)·3n=-(2n-2)·3n-2,∴Tn=

6、(n-1)·3n+1.3.解(1)设数列{an}的公差为d,由a2=1,S7=14,得a1+d=1,7a1+21d=14.解得a1=12,d=12,所以an=n2.∵b1·b2·b3·…·bn=2n2+n2=2n(n+1)2,∴b1·b2·b3·…·bn-1=2n(n-1)2(n≥2),两式相除,得bn=2n(n≥2).当n=1时,b1=2适合上式.∴bn=2n.(2)∵=bncos(anπ)=2ncosn2π,∴T2n=2cosπ2+22cosπ+23cos3π2+24cos(2π)+…+22n-1cos(2n-1)π2+22ncos(nπ)则T2n=22c

7、osπ+24cos(2π)+26cos(3π)+…+22ncos(nπ)=-22+24-26+…+(-1)n·22n=-4×[1-(-4)n]1+4=-4+(-4)n+15.4.解(1)由题意可得a1q3=9a1q,a1(1-q3)1-q=13,q>0,解得a1=1,q=3,所以an=3n-1,Sn=1-3n1-3=3n-12.(2)假设存在常数λ,使得数列{Sn+λ}是等比数列,因为S1+λ=λ+1,S2+λ=λ+4,S3+λ=λ+13,所以(λ+4)2=(λ+1)(λ+13),解得λ=12,此时Sn+12=12×3n,则Sn+1+12Sn+12=3,故存在

8、常数λ=12,使得数列Sn+12是等比