1、高考大题专项练三高考中的数列1.(2021新高考)已知数列an满足a1=1,an+1=an+1,n为奇数,an+2,n为偶数.(1)记bn=a2n,写出b1,b2,并求数列bn的通项公式;(2)求an的前20项和.2.(2021四川成都七中高三月考)设等差数列an的前n项和为Sn,已知a2=3,且S5=4a3+5.(1)求an的通项公式;(2)若1a1a2+1a2a3+1anan+11,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列bn满足b1=1,数列(bn+1-bn)an的前n项和为2n2+n.(1)求q的值;(2)求数列bn的通项公式.6.(2021北京东城区模拟)已知数
2、列an的前n项和是An,数列bn的前n项和是Bn,若a1=1,an+1=2an+1,nN*,再从三个条件:Bn=-n2+21n;Bn+1-bn=Bn-2,b1=20;bn=22-2log2(an+1)中任选一组作为已知条件,完成下面问题的解答.(1)求数列an,bn的通项公式;(2)定义:a*b=a,ab,b,ab,记cn=an*bn,求数列cn的前n项和Tn.7.已知正项数列an的首项a1=1,前n项和Sn满足an=Sn+Sn-1(n2).(1)求证:Sn为等差数列,并求数列an的通项公式;(2)记数列1anan+1的前n项和为Tn,若对任意的nN*,不等式4Tna2-a恒成立,求实数a的取
3、值范围.8.(2021天津高考)已知an是公差为2的等差数列,其前8项和为64.bn是公比大于0的等比数列,b1=4,b3-b2=48.(1)求an和bn的通项公式;(2)记cn=b2n+1bn,nN*,证明cn2-c2n是等比数列;证明k=1nakak+1ck2-c2k22(nN*).答案：1.解(1)b1=a2=a1+1=2,b2=a4=a3+1=a2+2+1=5.由bn+1=a2n+2=a2n+1+1=a2n+2+1=a2n+3,得bn+1-bn=a2n+3-a2n=3.所以bn是首项为2,公差为3的等差数列,所以bn=2+(n-1)3=3n-1.(2)由(1)知,数列an的偶数项组成的
4、数列是以3为公差的等差数列,由已知得an=an+1-1,n为奇数,所以数列an的奇数项组成的数列也是以3为公差的等差数列.设数列an的前n项和为Sn,则S20=(a1+a3+a5+a19)+(a2+a4+a20)=10+10923+20+10923=300,所以an的前20项和为300.2.解(1)设数列an的公差为d.法一:由题意,得a2=a1+d=3,5a1+10d=4(a1+2d)+5,解得a1=1,d=2.故an的通项公式为an=2n-1(nN*).法二:由S5=5(a1+a5)2=5a3,又S5=4a3+5,即a3=5.d=a3-a2=2,故an的通项公式为an=a2+(n-2)d=
5、2n-1(nN*).(2)设1anan+1的前n项和为Tn.1anan+1=1(2n-1)(2n+1)=1212n-1-12n+1,Tn=121-13+13-15+12n-1-12n+1=121-12n+112,由题设不等式恒成立,有1m-112,解得10,可得q=2,故bn=2n-1.所以,Tn=1-2n1-2=2n-1.设等差数列an的公差为d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,从而a1=1,d=1,故an=n.所以,Sn=n(n+1)2.(2)由(1),有T1+T2+Tn=(21+22+2n)-n=2(1-2n)1-2-n=2n+1-
6、n-2.由Sn+(T1+T2+Tn)=an+4bn可得,n(n+1)2+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍)或n=4.所以,n的值为4.5.解(1)由a4+2是a3,a5的等差中项,得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8q+1q=20,解得q=2或q=12,因为q1,所以q=2.(2)设cn=(bn+1-bn)an,数列cn前n项和为Sn,由Sn=2n2+n,cn=S1,n=1,Sn-Sn-1,n2,解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)12n-1
7、.故bn-bn-1=(4n-5)12n-2,n2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+(b3-b2)+(b2-b1)=(4n-5)12n-2+(4n-9)12n-3+712+3.设Tn=3+712+11122+(4n-5)12n-2,n2,12Tn=312+7122+(4n-9)12n-2+(4n-5)12n-1,所以12Tn=3+412+4122+412n-2-(4n-5)12n-1,因此Tn=14-(4n+3)12n-2,n2,又b1=1,所以bn=15-(4n+3)12n-2.6.解(1)由an+1=2an+1,得an+1+1=2(an+1),又a1=1,则a1+1=
8、2,数列an+1是以2为首项,2为公比的等比数列,an+1=2n,即an=2n-1.若选,当n=1时,b1=B1=20;当n2时,bn=Bn-Bn-1=22-2n.bn=22-2n.若选,由Bn+1-bn=Bn-2,得bn+1-bn=-2,所以数列bn是以20为首项,-2为公差的等差数列,bn=22-2n.若选,则bn=22-2log2(an+1)=22-2n.(2)由(1)知cn=an*bn=2n-1,1n3,22-2n,n4(nN*),当1n3时,Tn=2(1-2n)1-2-n=2n+1-2-n;当n4时,Tn=1+3+7+14+12+(22-2n)=-n2+21n-43.Tn=2n+1-
9、2-n,1n3,-n2+21n-43,n4(nN*).7.(1)证明因为an=Sn+Sn-1,n2,所以Sn-Sn-1=Sn+Sn-1,即Sn-Sn-1=1,所以数列Sn是首项为S1=a1=1,公差为1的等差数列,得Sn=n,所以an=Sn+Sn-1=n+(n-1)=2n-1(n2),当n=1时,a1=1也适合,所以an=2n-1.(2)解因为1anan+1=1(2n-1)(2n+1)=1212n-1-12n+1,所以Tn=121-13+13-15+12n-1-12n+1=121-12n+1.所以Tn12.要使不等式4Tn0),所以b3-b2=b1q2-b1q=4(q2-q)=48,解得q=4
10、(负值舍去),所以bn=b1qn-1=4n,nN*.(2)证明由题意,cn=b2n+1bn=42n+2-c2n=42n+14n2-44n+142n=24n,所以cn2-c2n0,且cn+12-c2n+2cn2-c2n=24n+124n=4,所以数列cn2-c2n是等比数列.由题意知,anan+1cn2-c2n=(2n-1)(2n+1)24n=4n2-1222n4n2222n,所以anan+1cn2-c2n4n2222n=2n22n=12n2n-1,所以k=1nakak+1ck2-c2k12k=1nk2k-1,设Tn=k=1nk2k-1=120+221+322+n2n-1,则12Tn=121+222+323+n2n,两式相减得12Tn=1+12+122+12n-1-n2n=11-12n1-12-n2n=2-n+22n,所以Tn=4-n+22n-1,所以k=1nakak+1ck2-c2k12k=1nk2k-1=124-n+22n-122.6
