天津高考数学二轮复习专题能力训练11等差数列与等比数列文.docx

专题能力训练11等差数列与等比数列一、能力突破训练1.已知等比数列an满足a1=,a3a5=4(a4-1),则a2=() A.2B.1C.D.答案：C解析：a3a5=4(a4-1),=4(a4-1),解得a4=2.又a4=a1q3,且a1=,q=2,a2=a1q=.2.在等差数列an中,a1+a2+a3=3,a18+a19+a20=87,则此数列前20项的和等于()A.290B.300C.580D.600答案：B解析：由a1+a2+a3=3,a18+a19+a20=87,得a1+a20=30,故S20=300.3.设an是等比数列,Sn是an的前n项和.对任意正整数n,有an+2an+1+an+2=0,又a1=2,则S101的值为()A.2B.200C.-2D.0答案：A解析：设公比为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.已知an是等差数列,公差d不为零,前n项和是Sn,若a3,a4,a8成等比数列,则()A.a1d0,dS40B.a1d0,dS40,dS40D.a1d0答案：B解析：设an的首项为a1,公差为d,则a3=a1+2d,a4=a1+3d,a8=a1+7d.a3,a4,a8成等比数列,(a1+3d)2=(a1+2d)(a1+7d),即3a1d+5d2=0.d0,a1d=-d20,且a1=-d.dS4=2d(2a1+3d)=-d20,解得a1=2,q=2,所以an=2n.(2)由题意知:S2n+1=(2n+1)bn+1,又S2n+1=bnbn+1,bn+10,所以bn=2n+1.令cn=,则cn=,因此Tn=c1+c2+cn=+.又Tn=+,两式相减得Tn=+-,所以Tn=5-.二、思维提升训练12.已知数列an,bn满足a1=b1=1,an+1-an=2,nN*,则数列的前10项的和为()A.(49-1)B.(410-1)C.(49-1)D.(410-1)答案：D解析：由a1=1,an+1-an=2,得an=2n-1.由=2,b1=1得bn=2n-1.则=22(n-1)=4n-1,故数列前10项和为=(410-1).13.若数列an为等比数列,且a1=1,q=2,则Tn=+等于()A.1-B.C.1-D.答案：B解析：因为an=12n-1=2n-1,所以anan+1=2n-12n=22n-1=24n-1,所以=.所以是等比数列.故Tn=+=.14.如图,点列An,Bn分别在某锐角的两边上,且|AnAn+1|=|An+1An+2|,AnAn+2,nN*,|BnBn+1|=|Bn+1Bn+2|,BnBn+2,nN*.(PQ表示点P与Q不重合)若dn=|AnBn|,Sn为AnBnBn+1的面积,则()A.Sn是等差数列B.是等差数列C.dn是等差数列D.是等差数列答案：A解析：如图,延长AnA1,BnB1交于P,过An作对边BnBn+1的垂线,其长度记为h1,过An+1作对边Bn+1Bn+2的垂线,其长度记为h2,则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).设此锐角为,则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 均为定值,Sn+1-Sn为定值.Sn是等差数列.故选A.15.已知等比数列an的首项为,公比为-,其前n项和为Sn,若ASn-B对nN*恒成立,则B-A的最小值为.答案：解析：易得Sn=1-,因为y=Sn-在区间上单调递增(y0),所以yA,B,因此B-A的最小值为-=.16.已知数列an的首项为1,Sn为数列an的前n项和,Sn+1=qSn+1,其中q0,nN*.(1)若a2,a3,a2+a3成等差数列,求数列an的通项公式;(2)设双曲线x2-=1的离心率为en,且e2=2,求+.解(1)由已知,Sn+1=qSn+1,Sn+2=qSn+1+1,两式相减得到an+2=qan+1,n1.又由S2=qS1+1得到a2=qa1,故an+1=qan对所有n1都成立.所以,数列an是首项为1,公比为q的等比数列.从而an=qn-1.由a2,a3,a2+a3成等差数列,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.所以双曲线x2-=1的离心率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.若数列an是公差为正数的等差数列,且对任意nN*有anSn=2n3-n2.(1)求数列an的通项公式.(2)是否存在数列bn,使得数列anbn的前n项和为An=5+(2n-3)2n-1(nN*)?若存在,求出数列bn的通项公式及其前n项和Tn;若不存在,请说明理由.解(1)设等差数列an的公差为d,则d0,an=dn+(a1-d),Sn=dn2+n.对任意nN*,恒有anSn=2n3-n2,则dn+(a1-d)=2n3-n2,即dn+(a1-d)=2n2-n.d0,an=2n-1.(2)数列anbn的前n项和为An=5+(2n-3)2n-1(nN*),当n=1时,a1b1=A1=4,b1=4,当n2时,anbn=An-An-1