廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 高考大題專項(xiàng)練三 高考中的數(shù)列 文.docx

高考大題專項(xiàng)練三高考中的數(shù)列1.(2018全國(guó),文17)等比數(shù)列an中,a1=1,a5=4a3.(1)求an的通項(xiàng)公式;(2)記Sn為an的前n項(xiàng)和,若Sm=63,求m.解(1)設(shè)an的公比為q,由題設(shè)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,則Sn=1-(-2)n3.由Sm=63得(-2)m=-188,此方程沒有正整數(shù)解.若an=2n-1,則Sn=2n-1.由Sm=63得2m=64,解得m=6.綜上,m=6.2.設(shè)數(shù)列an的前n項(xiàng)和為Sn,已知a1=3,Sn+1=3Sn+3.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn=nan+1-an,求數(shù)列bn的前n項(xiàng)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.當(dāng)n2時(shí),an=Sn-Sn-1=3n+12-3n2=3n,a1也適合.an=3n.(方法二)由Sn+1=3Sn+3(nN*),可知當(dāng)n2時(shí),Sn=3Sn-1+3,兩式相減,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.已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=2an-1;數(shù)列bn滿足bn-1-bn=bnbn-1(n2,nN*),b1=1.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)求數(shù)列anbn的前n項(xiàng)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),兩式相減,得Sn-Sn-1=2an-2an-1,即an=2an-2an-1.故an=2an-1,n2.所以數(shù)列an是首項(xiàng)為1,公比為2的等比數(shù)列.故an=12n-1=2n-1.由bn-1-bn=bnbn-1(n2,nN*),得1bn-1bn-1=1.又b1=1,數(shù)列1bn是首項(xiàng)為1,公差為1的等差數(shù)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.兩式相減,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)設(shè)an是等差數(shù)列,其前n項(xiàng)和為Sn(nN*);bn是等比數(shù)列,公比大于0,其前n項(xiàng)和為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,求正整數(shù)n的值.解(1)設(shè)等比數(shù)列bn的公比為q.由b1=1,b3=b2+2,可得q2-q-2=0.因?yàn)閝>0,可得q=2,故bn=2n-1.所以,Tn=1-2n1-2=2n-1.設(shè)等差數(shù)列an的公差為d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,從而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的值為4.5.已知f(x)=2sin2x,集合M=x|f(x)|=2,x>0,把M中的元素從小到大依次排成一列,得到數(shù)列an,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)記bn=1an+12,設(shè)數(shù)列bn的前n項(xiàng)和為Tn,求證:Tn<14.(1)解f(x)=2sin2x,集合M=x|f(x)|=2,x>0,則2x=k+2,解得x=2k+1(kZ),把M中的元素從小到大依次排成一列,得到數(shù)列an,所以an=2n-1.(2)證明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)已知等比數(shù)列an的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列bn滿足b1=1,數(shù)列(bn+1-bn)an的前n項(xiàng)和為2n2+n.(1)求q的值;(2)求數(shù)列bn的通項(xiàng)公式.解(1)由a4+2是a3,a5的等差中項(xiàng),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?yàn)閝>1,所以q=2.(2)設(shè)cn=(bn+1-bn)an,數(shù)列cn前n項(xiàng)和為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.設(shè)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.已知正項(xiàng)數(shù)列an的首項(xiàng)a1=1,前n項(xiàng)和Sn滿足an=Sn+Sn-1(n2).(1)求證:Sn為等差數(shù)列,并求數(shù)列an的通項(xiàng)公式;(2)記數(shù)列1anan+1的前n項(xiàng)和為Tn,若對(duì)任意的nN*,不等式4Tn<a2-a恒成立,求實(shí)數(shù)a的取值范圍.解(1)因?yàn)閍n=Sn+Sn-1,所以Sn-Sn-1=Sn+Sn-1,即Sn-Sn-1=1,所以數(shù)列Sn是首項(xiàng)為S1=a1=1,公差為1的等差數(shù)列,得Sn=n,所以an=Sn+Sn-1=n+(n-1)=2n-1(n2),當(dāng)n=1時(shí),a1=1也適合,所以an=2n-1.(2)因?yàn)?anan+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,故實(shí)數(shù)a的取值范圍是(-,-12,+).8.已知數(shù)列an是公比為12的等比數(shù)列,其前n項(xiàng)和為Sn,且1-a2是a1與1+a3的等比中項(xiàng),數(shù)列bn是等差數(shù)列,其前n項(xiàng)和Tn滿足Tn=nbn+1(為常數(shù),且1),其中b1=8.(1)求數(shù)列an的通項(xiàng)公式及的值;(2)比較1T1+1T2+1T3+1Tn與12Sn的大小.解(1)由題意,得(1-a2)2=a1(a3+1),即1-12a12=a114a1+1,解得a1=12.故an=12n.設(shè)等差數(shù)列bn的公差為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,則12Sn=12-12n+114.由(1)知Tn=12nbn+1,當(dāng)n=1時(shí),T1=b1=12b2,即b2=2b1=16,故公差d=b2-b1=8,則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.