广西2020版高考数学一轮复习 高考大题专项练三 高考中的数列 文.docx

高考大题专项练三高考中的数列1.(2018全国,文17)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.设数列an的前n项和为Sn,已知a1=3,Sn+1=3Sn+3.(1)求数列an的通项公式;(2)若bn=nan+1-an,求数列bn的前n项和Tn.解(1)(方法一)Sn+1=3Sn+3,Sn+1+32=3Sn+32.Sn+32=S1+323n-1=923n-1=3n+12.当n2时,an=Sn-Sn-1=3n+12-3n2=3n,a1也适合.an=3n.(方法二)由Sn+1=3Sn+3(nN*),可知当n2时,Sn=3Sn-1+3,两式相减,得an+1=3an(n2).又a1=3,代入Sn+1=3Sn+3,得a2=9,故an=3n.(2)bn=nan+1-an=n3n+1-3n=12n3n,Tn=1213+232+333+n3n,13Tn=12132+233+334+n-13n+n3n+1,由-,得23Tn=1213+132+133+134+13n-n3n+1,解得Tn=38-2n+383n.3.已知数列an的前n项和为Sn,且Sn=2an-1;数列bn满足bn-1-bn=bnbn-1(n2,nN*),b1=1.(1)求数列an,bn的通项公式;(2)求数列anbn的前n项和Tn.解(1)由Sn=2an-1,得S1=a1=2a1-1,故a1=1.又Sn=2an-1,Sn-1=2an-1-1(n2),两式相减,得Sn-Sn-1=2an-2an-1,即an=2an-2an-1.故an=2an-1,n2.所以数列an是首项为1,公比为2的等比数列.故an=12n-1=2n-1.由bn-1-bn=bnbn-1(n2,nN*),得1bn-1bn-1=1.又b1=1,数列1bn是首项为1,公差为1的等差数列.1bn=1+(n-1)1=n.bn=1n.(2)由(1)得anbn=n2n-1.Tn=120+221+n2n-1,2Tn=121+222+n2n.两式相减,得-Tn=1+21+2n-1-n2n=1-2n1-2-n2n=-1+2n-n2n.Tn=(n-1)2n+1.4.(2018天津,文18)设an是等差数列,其前n项和为Sn(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q>0,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-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f(x)=2sin2x,集合M=x|f(x)|=2,x>0,把M中的元素从小到大依次排成一列,得到数列an,nN*.(1)求数列an的通项公式;(2)记bn=1an+12,设数列bn的前n项和为Tn,求证:Tn<14.(1)解f(x)=2sin2x,集合M=x|f(x)|=2,x>0,则2x=k+2,解得x=2k+1(kZ),把M中的元素从小到大依次排成一列,得到数列an,所以an=2n-1.(2)证明bn=1an+12=1(2n+1)2<14n2+4n=141n-1n+1,故Tn=b1+b2+bn<141-12+12-13+1n-1n+1=141-1n+1<14.6.(2018浙江,20)已知等比数列an的公比q>1,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=20,解得q=2或q=12,因为q>1,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+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-2.7.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4Tn<a2-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,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Tn<12.要使不等式4Tn<a2-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+1<14.由可知1T1+1T2+1Tn<12Sn.