2021高考数学（理）导学大一轮人教A广西专用高考大题专项练三　高考中的数列 WORD版含解析.docx

1、高考大题专项练三高考中的数列高考大题专项练第6页1.在等比数列an中,a1=1,a5=4a3.(1)求an的通项公式;(2)记Sn为an的前n项和,若Sm=63,求m.解:(1)设an的公比为q,由题设得an=qn-1.由已知得q4=4q2,解得q=0(舍去),q=-2或q=2.故an=(-2)n-1或an=2n-1.(2)若an=(-2)n-1,则Sn=1-(-2)n3.由Sm=63得(-2)m=-188,此方程没有正整数解.若an=2n-1,则Sn=2n-1.由Sm=63得2m=64,解得m=6.综上,m=6.2.已知等差数列an的前n项和为Sn,a3=5,S5=3S3-2.(1)求an的

2、通项公式;(2)设bn=2an,求数列bn的前n项和Tn.解:(1)设等差数列an的公差为d,a3=5,S5=3S3-2,a3=a1+2d=5,5a1+10d=3(3a1+3d)-2,a1=1,d=2,an=2n-1.(2)bn=2an=22n-1,bn+1bn=22(n+1)-122n-1=22n+122n-1=22=4,b1=2.数列bn是等比数列,公比为4,首项为2.Tn=2(1-4n)1-4=23(4n-1).3.(2019全国,理19)已知数列an和bn满足a1=1,b1=0,4an+1=3an-bn+4,4bn+1=3bn-an-4.(1)证明:an+bn是等比数列,an-bn是等

3、差数列;(2)求an和bn的通项公式.(1)证明由题设得4(an+1+bn+1)=2(an+bn),即an+1+bn+1=12(an+bn).又因为a1+b1=1,所以an+bn是首项为1,公比为12的等比数列.由题设得4(an+1-bn+1)=4(an-bn)+8,即an+1-bn+1=an-bn+2.又因为a1-b1=1,所以an-bn是首项为1,公差为2的等差数列.(2)解由(1)知,an+bn=12n-1,an-bn=2n-1,所以an=12(an+bn)+(an-bn)=12n+n-12,bn=12(an+bn)-(an-bn)=12n-n+12.4.设an是等差数列,其前n项和为S

4、n(nN*);bn是等比数列,公比大于0,其前n项和为Tn(nN*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+Tn)=an+4bn,求正整数n的值.解:(1)设等比数列bn的公比为q.由b1=1,b3=b2+2,可得q2-q-2=0.因为q0,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=

5、(21+22+2n)-n=2(1-2n)1-2-n=2n+1-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.已知等比数列an的公比q1,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.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

6、=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.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

7、+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.设Sn为等差数列an的前n项和,已知S3=a7,a8-2a3=3.(1)求an;(2)设bn=1Sn,数列bn的前n项和为Tn,求证:Tn34-1n+1(nN*).(1)解设等差数列an的公差为d,由题意,得3a1+3d=a1+6d,(a1+7d)-2(a1+2d)=3,解得a1=3,d=2.故an=a1+(n-1)d=2n+1.(2)证明a1=3,d=2,Sn=na1+n(n-1)2d=n(n+2).bn=1n(n+2)=121n-1

8、n+2.Tn=b1+b2+bn-1+bn=121-13+12-14+1n-1-1n+1+1n-1n+2=121+12-1n+1-1n+2121+12-1n+1-1n+1=34-1n+1,故Tn34-1n+1.7.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的取值范围.解:(1)因为an=Sn+Sn-1,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

9、,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)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4Tna2-a恒成立,只需2a2-a恒成立,解得a-1或a2,故实数a的取值范围是(-,-12,+).8.已知数列an是公比为12的等比数列,其前n项和为Sn,且1-a2是a1与1+a3的等比中项,数列bn是等差数列,其前n项和Tn满足Tn=nbn+1(为常数,且1),其中b1=8.(1)求

10、数列an的通项公式及的值;(2)比较1T1+1T2+1T3+1Tn与12Sn的大小.解:(1)由题意,得(1-a2)2=a1(a3+1),即1-12a12=a114a1+1,解得a1=12.故an=12n.设等差数列bn的公差为d,又T1=b2,T2=2b3,即8=(8+d),16+d=2(8+2d),解得=12,d=8或=1,d=0(舍去),故=12.(2)由(1)知Sn=1-12n,则12Sn=12-12n+114.由(1)知Tn=12nbn+1,当n=1时,T1=b1=12b2,即b2=2b1=16,故公差d=b2-b1=8,则bn=8n,又Tn=nbn+1,故Tn=4n2+4n,即1Tn=14n(n+1)=141n-1n+1.因此,1T1+1T2+1Tn=141-12+12-13+1n-1n+1=141-1n+114.由可知1T1+1T2+1Tn12Sn.