2022高考数学一轮复习-高考大题专项数列北师大版.docx

2022高考数学一轮复习 高考大题专项数列北师大版

2022高考数学一轮复习 高考大题专项数列北师大版

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高考大题专项(三)　数列
1.(2020广东天河区模拟)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湖南郴州二模,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cn=an·bn,求数列{cn}的前n项和Tn.

3.(2020安徽合肥4月质检二,理17)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{cn}满足cn=bncos(anπ),求数列{cn}的前2n项和T2n.

4.(2020山西长治二模)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天津,19)已知{an}为等差数列,{bn}为等比数列,a1=b1=1,a5=5(a4-a3),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+2<Sn+12(n∈N+);
(3)对任意的正整数n,设cn=(3an-2)bnanan+2,n为奇数,an-1bn+1,n为偶数.求数列{cn}的前2n项和.

参考答案

高考大题专项(三)　数列
1.解(1)当n=1时,6a1=a12+3a1+2,且a1<2,解得a1=1.
当n≥2时,6an=6Sn-6Sn-1=an2+3an+2-(an-12+3an-1+2).
化简,得(an+an-1)(an-an-1-3)=0,因为an>0,所以an-an-1=3,
所以数列{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)=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,cn=an·bn=(2n-1)·3n-1,
∴Tn=c1+c2+c3+…+cn=1×1+3×3+5×32+…+(2n-1)·3n-1,①
3Tn=1×3+3×32+…+(2n-3)·3n-1+(2n-1)·3n,②
①-②,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=(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)∵cn=bncos(anπ)=2ncosn2π,
∴T2n=2cosπ2+22cosπ+23cos3π2+24cos(2π)+…+22n-1cos(2n-1)π2+22ncos(nπ)
则T2n=22cosπ+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,故存在常数λ=12,使得数列Sn+12是等比数列.
5.(1)解设等差数列{an}的公差为d,等比数列{bn}的公比为q.由a1=1,a5=5(a4-a3),可得d=1,从而{an}的通项公式为an=n.由b1=1,b5=4(b4-b3),又q≠0,可得q2-4q+4=0,解得q=2,从而{bn}的通项公式为bn=2n-1.
(2)证明由(1)可得Sn=n(n+1)2,故SnSn+2=14n(n+1)(n+2)(n+3),Sn+12=14(n+1)2(n+2)2,从而SnSn+2-Sn+12=-12(n+1)(n+2)<0,所以SnSn+2<Sn+12.
(3)解当n为奇数时,cn=(3an-2)bnanan+2=(3n-2)2n-1n(n+2)=2n+1n+2-2n-1n;当n为偶数时,cn=an-1bn+1=n-12n.对任意的正整数n,有∑k=1nc2k-1=∑k=1n22k2k+1-22k-22k-1=22n2n+1-1,
∑k=1nc2k=∑k=1n2k-14k=14+342+543+…+2n-14n.①
由①得14∑k=1nc2k=142+343+…+2n-34n+2n-14n+1.②
由①②得34∑k=1nc2k=14+242+…+24n-2n-14n+1=241-14n1-14-14-2n-14n+1,从而得∑k=1nc2k=59-6n+59×4n.
因此,∑k=12nck=∑k=1nc2k-1+∑k=1nc2k=4n2n+1-6n+59×4n-49.所以,数列{cn}的前2n项和为4n2n+1-6n+59×4n-49.