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年级：姓名：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a10,6Sn=an2+3an+2,nN+.(1)求数列an的通项公式;(2)若任意nN+,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(d1),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=anbn,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b1b2b3bn=2n2+n2.(1)求数列an和bn的通项公式;(2)若数列cn满足cn=bncos(an),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q0.(1)求an及Sn.(2)是否存在常数,使得数列Sn+是等比数列?若存在,求出的值;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+2Sn+12(nN+);(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a10,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=36n(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,nN+.b1=a1=1,q=32d=322=3,数列bn的通项公式为bn=13n-1=3n-1,nN+.(2)由(1)知,cn=anbn=(2n-1)3n-1,Tn=c1+c2+c3+cn=11+33+532+(2n-1)3n-1,3Tn=13+332+(2n-3)3n-1+(2n-1)3n,-,得-2Tn=1+23+232+23n-1-(2n-1)3n=1+2(3+32+3n-1)-(2n-1)3n=1+23-3n1-3-(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.b1b2b3bn=2n2+n2=2n(n+1)2,b1b2b3bn-1=2n(n-1)2(n2),两式相除,得bn=2n(n2).当n=1时,b1=2适合上式.bn=2n.(2)cn=bncos(an)=2ncosn2,T2n=2cos2+22cos+23cos32+24cos(2)+22n-1cos(2n-1)2+22ncos(n)则T2n=22cos+24cos(2)+26cos(3)+22ncos(n)=-22+24-26+(-1)n22n=-41-(-4)n1+4=-4+(-4)n+15.4.解(1)由题意可得a1q3=9a1q,a1(1-q3)1-q=13,q0,解得a1=1,q=3,所以an=3n-1,Sn=1-3n1-3=3n-12.(2)假设存在常数,使得数列Sn+是等比数列,因为S1+=+1,S2+=+4,S3+=+13,所以(+4)2=(+1)(+13),解得=12,此时Sn+12=123n,则Sn+1+12Sn+12=3,故存在常数=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q0,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+2Sn+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.由得14k=1nc2k=142+343+2n-34n+2n-14n+1.由得34k=1nc2k=14+242+24n-2n-14n+1=241-14n1-14-14-2n-14n+1,从而得k=1nc2k=59-6n+594n.因此,k=12nck=k=1nc2k-1+k=1nc2k=4n2n+1-6n+594n-49.所以,数列cn的前2n项和为4n2n+1-6n+594n-49.