2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題四 數(shù)列 專題能力訓(xùn)練11 等差數(shù)列與等比數(shù)列 文1.已知等比數(shù)列an滿足a1=,a3a5=4(a4-1),則a2=()A.2B.1C.D.2.在等差數(shù)列an中,a1+a2+a3=3,a18+a19+a20=87,則此數(shù)列前20項(xiàng)的和等于()A.290B.300C.580D.6003.設(shè)an是等比數(shù)列,Sn是an的前n項(xiàng)和.對(duì)任意正整數(shù)n,有an+2an+1+an+2=0,又a1=2,則S101的值為()A.2B.200C.-2D.04.已知an是等差數(shù)列,公差d不為零,前n項(xiàng)和是Sn,若a3,a4,a8成等比數(shù)列,則()A.a1d>0,dS4>0B.a1d<0,dS4<0C.a1d>0,dS4<0D.a1d<0,dS4>05.在等比數(shù)列an中,滿足a1+a2+a3+a4+a5=3,=15,則a1-a2+a3-a4+a5的值是()A.3B.C.-D.56.在數(shù)列an中,a1=2,an+1=2an,Sn為an的前n項(xiàng)和.若Sn=126,則n=. 7.已知等比數(shù)列an為遞增數(shù)列,且=a10,2(an+an+2)=5an+1,則數(shù)列的通項(xiàng)公式an=. 8.設(shè)x,y,z是實(shí)數(shù),若9x,12y,15z成等比數(shù)列,且成等差數(shù)列,則=. 9.(2018全國(guó),文17)在等比數(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.10.已知等差數(shù)列an和等比數(shù)列bn滿足a1=b1=1,a2+a4=10,b2b4=a5.(1)求an的通項(xiàng)公式;(2)求和:b1+b3+b5+b2n-1.11.設(shè)數(shù)列an滿足a1+3a2+(2n-1)an=2n.(1)求an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和.二、思維提升訓(xùn)練12.已知數(shù)列an,bn滿足a1=b1=1,an+1-an=2,nN*,則數(shù)列的前10項(xiàng)的和為()A. (49-1)B. (410-1)C. (49-1)D. (410-1)13.若數(shù)列an為等比數(shù)列,且a1=1,q=2,則Tn=+等于()A.1-B.C.1-D.14.如圖,點(diǎn)列An,Bn分別在某銳角的兩邊上,且|AnAn+1|=|An+1An+2|,AnAn+2,nN*,|BnBn+1|=|Bn+1Bn+2|,BnBn+2,nN*.(PQ表示點(diǎn)P與Q不重合)若dn=|AnBn|,Sn為AnBnBn+1的面積,則()A.Sn是等差數(shù)列B.是等差數(shù)列C.dn是等差數(shù)列D.是等差數(shù)列15.已知等比數(shù)列an的首項(xiàng)為,公比為-,其前n項(xiàng)和為Sn,若ASn-B對(duì)nN*恒成立,則B-A的最小值為. 16.已知數(shù)列an的首項(xiàng)為1,Sn為數(shù)列an的前n項(xiàng)和,Sn+1=qSn+1,其中q>0,nN*.(1)若a2,a3,a2+a3成等差數(shù)列,求數(shù)列an的通項(xiàng)公式;(2)設(shè)雙曲線x2-=1的離心率為en,且e2=2,求+.17.若數(shù)列an是公差為正數(shù)的等差數(shù)列,且對(duì)任意nN*有an·Sn=2n3-n2.(1)求數(shù)列an的通項(xiàng)公式.(2)是否存在數(shù)列bn,使得數(shù)列anbn的前n項(xiàng)和為An=5+(2n-3)2n-1(nN*)?若存在,求出數(shù)列bn的通項(xiàng)公式及其前n項(xiàng)和Tn;若不存在,請(qǐng)說(shuō)明理由.專題能力訓(xùn)練11等差數(shù)列與等比數(shù)列一、能力突破訓(xùn)練1.C解析 a3a5=4(a4-1),=4(a4-1),解得a4=2.又a4=a1q3,且a1=,q=2,a2=a1q=.2.B解析 由a1+a2+a3=3,a18+a19+a20=87,得a1+a20=30,故S20=300.3.A解析 設(shè)公比為q,an+2an+1+an+2=0,a1+2a2+a3=0,a1+2a1q+a1q2=0,q2+2q+1=0,q=-1.又a1=2,S101=2.4.B解析 設(shè)an的首項(xiàng)為a1,公差為d,則a3=a1+2d,a4=a1+3d,a8=a1+7d.a3,a4,a8成等比數(shù)列,(a1+3d)2=(a1+2d)(a1+7d),即3a1d+5d2=0.d0,a1d=-d2<0,且a1=-d.dS4=2d(2a1+3d)=-d2<0,故選B.5.D解析 由條件知=5,故a1-a2+a3-a4+a5=5.6.6解析 an+1=2an,即=2,an是以2為公比的等比數(shù)列.又a1=2,Sn=126.2n=64,n=6.7.2n解析 =a10,(a1q4)2=a1q9,a1=q,an=qn.2(an+an+2)=5an+1,2an(1+q2)=5anq,2(1+q2)=5q,解得q=2或q=(舍去),an=2n.8.解析 由題意知解得xz=y2=y2,x+z=y,從而-2=-2=.9.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=.由Sm=63得(-2)m=-188,此方程沒(méi)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=6.10.解 (1)設(shè)等差數(shù)列an的公差為d.因?yàn)閍2+a4=10,所以2a1+4d=10.解得d=2.所以an=2n-1.(2)設(shè)等比數(shù)列bn的公比為q.因?yàn)閎2b4=a5,所以b1qb1q3=9.解得q2=3.所以b2n-1=b1q2n-2=3n-1.從而b1+b3+b5+b2n-1=1+3+32+3n-1=.11.解 (1)因?yàn)閍1+3a2+(2n-1)an=2n,故當(dāng)n2時(shí),a1+3a2+(2n-3)an-1=2(n-1).兩式相減得(2n-1)an=2.所以an=(n2).又由題設(shè)可得a1=2,從而an的通項(xiàng)公式為an=.(2)記的前n項(xiàng)和為Sn.由(1)知,則Sn=+.二、思維提升訓(xùn)練12.D解析 由a1=1,an+1-an=2,得an=2n-1.由=2,b1=1得bn=2n-1.則=22(n-1)=4n-1,故數(shù)列前10項(xiàng)和為(410-1).13.B解析 因?yàn)閍n=1×2n-1=2n-1,所以anan+1=2n-1·2n=22n-1=2×4n-1,所以.所以是等比數(shù)列.故Tn=+.14.A解析 如圖,延長(zhǎng)AnA1,BnB1交于P,過(guò)An作對(duì)邊BnBn+1的垂線,其長(zhǎng)度記為h1,過(guò)An+1作對(duì)邊Bn+1Bn+2的垂線,其長(zhǎng)度記為h2,則Sn=|BnBn+1|×h1,Sn+1=|Bn+1Bn+2|×h2.Sn+1-Sn=|Bn+1Bn+2|h2-|BnBn+1|h1.|BnBn+1|=|Bn+1Bn+2|,Sn+1-Sn=|BnBn+1|(h2-h1).設(shè)此銳角為,則h2=|PAn+1|sin ,h1=|PAn|sin ,h2-h1=sin (|PAn+1|-|PAn|)=|AnAn+1|sin .Sn+1-Sn=|BnBn+1|AnAn+1|sin .|BnBn+1|,|AnAn+1|,sin 均為定值,Sn+1-Sn為定值.Sn是等差數(shù)列.故選A.15.解析 易得Sn=1-,因?yàn)閥=Sn-在區(qū)間上單調(diào)遞增(y0),所以yA,B,因此B-A的最小值為.16.解 (1)由已知,Sn+1=qSn+1,Sn+2=qSn+1+1,兩式相減得到an+2=qan+1,n1.又由S2=qS1+1得到a2=qa1,故an+1=qan對(duì)所有n1都成立.所以,數(shù)列an是首項(xiàng)為1,公比為q的等比數(shù)列.從而an=qn-1.由a2,a3,a2+a3成等差數(shù)列,可得2a3=a2+a2+a3.所以a3=2a2,故q=2.所以an=2n-1(nN*).(2)由(1)可知,an=qn-1.所以雙曲線x2-=1的離心率en=.由e2=2,解得q=.所以+=(1+1)+(1+q2)+1+q2(n-1)=n+1+q2+q2(n-1)=n+=n+(3n-1).17.解 (1)設(shè)等差數(shù)列an的公差為d,則d>0,an=dn+(a1-d),Sn=dn2+n.對(duì)任意nN*,恒有an·Sn=2n3-n2,則dn+(a1-d)·=2n3-n2,即dn+(a1-d)·=2n2-n.d>0,an=2n-1.(2)數(shù)列anbn的前n項(xiàng)和為An=5+(2n-3)·2n-1(nN*),當(dāng)n=1時(shí),a1b1=A1=4,b1=4,當(dāng)n2時(shí),anbn=An-An-1=5+(2n-3)2n-1-5+(2n-5)2n-2=(2n-1)2n-2.bn=2n-2.假設(shè)存在數(shù)列bn滿足題設(shè),且數(shù)列bn的通項(xiàng)公式bn=T1=4,當(dāng)n2時(shí),Tn=4+=2n-1+3,當(dāng)n=1時(shí)也適合,數(shù)列bn的前n項(xiàng)和為Tn=2n-1+3.