高三理科數(shù)學(xué) 新課標(biāo)二輪復(fù)習(xí)專(zhuān)題整合高頻突破習(xí)題：專(zhuān)題四 數(shù)列 專(zhuān)題能力訓(xùn)練11 Word版含答案

專(zhuān)題能力訓(xùn)練11等差數(shù)列與等比數(shù)列能力突破訓(xùn)練1.(20xx甘肅肅南?？?在等差數(shù)列an中,a4+a10+a16=30,則a18-2a14的值為()A.20B.-20C.10D.-102.(20xx貴州黔東南州模擬)在各項(xiàng)均為正數(shù)的等比數(shù)列an中,若log2(a2a3a5a7a8)=5,則a1a9=()A.4B.5C.2D.253.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.(20xx廣西南寧適應(yīng)性測(cè)試)已知數(shù)列an滿(mǎn)足an+1an+1+1=12,且a2=2,則a4等于()A.-12B.23C.12D.116.(20xx三湘名校聯(lián)盟聯(lián)考三)已知各項(xiàng)均為正數(shù)的等差數(shù)列an的前n項(xiàng)和為Sn,S10=40,則a3a8的最大值為.7.設(shè)等比數(shù)列an滿(mǎn)足a1+a3=10,a2+a4=5,則a1a2an的最大值為.8.設(shè)x,y,z是實(shí)數(shù),若9x,12y,15z成等比數(shù)列,且1x,1y,1z成等差數(shù)列,則xz+zx=.9.已知Sn為數(shù)列an的前n項(xiàng)和,且a2+S2=31,an+1=3an-2n(nN*).(1)求證:an-2n為等比數(shù)列;(2)求數(shù)列an的前n項(xiàng)和Sn.10.已知數(shù)列an的前n項(xiàng)和為Sn,a1=1,an0,anan+1=Sn-1,其中為常數(shù).(1)證明:an+2-an=;(2)是否存在,使得an為等差數(shù)列?并說(shuō)明理由.11.已知數(shù)列an是等比數(shù)列.設(shè)a2=2,a5=16.(1)若a1+a2+a2n=t(a12+a22+an2),nN*,求實(shí)數(shù)t的值;(2)若在1a1與1a4之間插入k個(gè)數(shù)b1,b2,bk,使得1a1,b1,b2,bk,1a4,1a5成等差數(shù)列,求k的值.思維提升訓(xùn)練12.(20xx全國(guó),理12)幾位大學(xué)生響應(yīng)國(guó)家的創(chuàng)業(yè)號(hào)召,開(kāi)發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的興趣,他們推出了“解數(shù)學(xué)題獲取軟件激活碼”的活動(dòng).這款軟件的激活碼為下面數(shù)學(xué)問(wèn)題的答案:已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,其中第一項(xiàng)是20,接下來(lái)的兩項(xiàng)是20,21,再接下來(lái)的三項(xiàng)是20,21,22,依此類(lèi)推.求滿(mǎn)足如下條件的最小整數(shù)N:N>100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()A.440B.330C.220D.11013.若數(shù)列an為等比數(shù)列,且a1=1,q=2,則Tn=1a1a2+1a2a3+1anan+1等于()A.1-14nB.231-14nC.1-12nD.231-12n14.已知等比數(shù)列an的首項(xiàng)為43,公比為-13,其前n項(xiàng)和為Sn,若ASn-1SnB對(duì)nN*恒成立,則B-A的最小值為.15.無(wú)窮數(shù)列an由k個(gè)不同的數(shù)組成,Sn為an的前n項(xiàng)和,若對(duì)任意nN*,Sn2,3,則k的最大值為.16.等比數(shù)列an的各項(xiàng)均為正數(shù),且2a1+3a2=1,a32=9a2a6.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=log3a1+log3a2+log3an,求數(shù)列1bn的前n項(xiàng)和.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anSn=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.參考答案專(zhuān)題能力訓(xùn)練11等差數(shù)列與等比數(shù)列能力突破訓(xùn)練1.D解析對(duì)題目中下標(biāo)數(shù)值仔細(xì)觀察,有a4+a10+a16=303a10=30,a10=10,所以a18-2a14=-a10=-10.故選D.2.A解析由題意得log2(a2a3a5a7a8)=log2a55=5log2a5=5,所以a5=2.所以a1a9=a52=4.故選A.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=a1(1-q101)1-q=21-(-1)1011+1=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=-53d2<0,且a1=-53d.dS4=4d(a1+a4)2=2d(2a1+3d)=-23d2<0,故選B.5.D解析由已知得an+1+1an+1=2,則an+1是公比為2的等比數(shù)列,所以a4+1=(a2+1)22=12.所以a4=11.故選D.6.16解析S10=10(a1+a10)2=40a1+a10=a3+a8=8,a3a8a3+a822=822=16,當(dāng)且僅當(dāng)a3=a8=4時(shí)取等號(hào).7.64解析由已知a1+a3=10,a2+a4=a1q+a3q=5,兩式相除得a1+a3q(a1+a3)=105,解得q=12,a1=8,所以a1a2an=8n121+2+(n-1)=2-12n2+7n2,拋物線(xiàn)f(n)=-12n2+72n的對(duì)稱(chēng)軸為n=-722-12=3.5,又nN*,所以當(dāng)n=3或4時(shí),a1a2an取最大值為2-1232+732=26=64.8.3415解析由題意知(12y)2=9x15z,2y=1x+1z,解得xz=122915y2=1615y2,x+z=3215y,從而xz+zx=x2+z2xz=(x+z)2-2xzxz=(x+z)2xz-2=32152y21615y2-2=3415.9.(1)證明由an+1=3an-2n可得an+1-2n+1=3an-2n-2n+1=3an-32n=3(an-2n).又a2=3a1-2,則S2=a1+a2=4a1-2,得a2+S2=7a1-4=31,得a1=5,則a1-21=30.故an-2n為等比數(shù)列.(2)解由(1)可知an-2n=3n-1(a1-2)=3n,an=2n+3n,Sn=2(1-2n)1-2+3(1-3n)1-3=2n+1+3n+12-72.10.(1)證明由題設(shè),anan+1=Sn-1,an+1an+2=Sn+1-1,兩式相減,得an+1(an+2-an)=an+1.因?yàn)閍n+10,所以an+2-an=.(2)解由a1=1,a1a2=S1-1,得a2=-1.由(1)知,a3=+1.令2a2=a1+a3,解得=4.故an+2-an=4.由此可得a2n-1是首項(xiàng)為1,公差為4的等差數(shù)列,a2n-1=4n-3;a2n是首項(xiàng)為3,公差為4的等差數(shù)列,a2n=4n-1.所以an=2n-1,an+1-an=2.因此存在=4,使得數(shù)列an為等差數(shù)列.11.解設(shè)等比數(shù)列an的公比為q,由a2=2,a5=16,得q=2,a1=1.(1)a1+a2+a2n=t(a12+a22+an2),a1(1-q2n)1-q=ta12(1-q2n)1-q2,即1-22n1-2=t1-22n1-4對(duì)nN*都成立,t=3.(2)1a1=1,1a4=18,1a5=116,且1a1,b1,b2,bk,1a4,1a5成等差數(shù)列,公差d=1a5-1a4=-116,且1a4-1a1=(k+1)d,即18-1=(k+1)-116,解得k=13.思維提升訓(xùn)練12.A解析設(shè)數(shù)列的首項(xiàng)為第1組,接下來(lái)兩項(xiàng)為第2組,再接下來(lái)三項(xiàng)為第3組,以此類(lèi)推,設(shè)第n組的項(xiàng)數(shù)為n,則前n組的項(xiàng)數(shù)和為n(1+n)2.第n組的和為1-2n1-2=2n-1,前n組總共的和為2(1-2n)1-2-n=2n+1-2-n.由題意,N>100,令n(1+n)2>100,得n14且nN*,即N出現(xiàn)在第13組之后.若要使最小整數(shù)N滿(mǎn)足:N>100且前N項(xiàng)和為2的整數(shù)冪,則SN-Sn(1+n)2應(yīng)與-2-n互為相反數(shù),即2k-1=2+n(kN*,n14),所以k=log2(n+3),解得n=29,k=5.所以N=29(1+29)2+5=440,故選A.13.B解析因?yàn)閍n=12n-1=2n-1,所以anan+1=2n-12n=22n-1=24n-1,所以1anan+1=1214n-1.所以1anan+1是等比數(shù)列.故Tn=1a1a2+1a2a3+1anan+1=1211-14n1-14=231-14n.14.5972解析易得Sn=1-13n89,11,43,因?yàn)閥=Sn-1Sn在89,43上單調(diào)遞增(y0),所以y-1772,712A,B,因此B-A的最小值為712-1772=5972.15.4解析要滿(mǎn)足數(shù)列中的條件,涉及最多的項(xiàng)的數(shù)列可以為2,1,-1,0,0,0,所以最多由4個(gè)不同的數(shù)組成.16.解(1)設(shè)數(shù)列an的公比為q.由a32=9a2a6得a32=9a42,所以q2=19.由條件可知q>0,故q=13.由2a1+3a2=1得2a1+3a1q=1,所以a1=13.故數(shù)列an的通項(xiàng)公式為an=13n.(2)bn=log3a1+log3a2+log3an=-(1+2+n)=-n(n+1)2.故1bn=-2n(n+1)=-21n-1n+1,1b1+1b2+1bn=-21-12+12-13+1n-1n+1=-2nn+1.所以數(shù)列1bn的前n項(xiàng)和為-2nn+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=12dn2+a1-12dn.對(duì)任意nN*,恒有anSn=2n3-n2,則dn+(a1-d)12dn2+a1-12dn=2n3-n2,即dn+(a1-d)12dn+a1-12d=2n2-n.12d2=2,12d(a1-d)+da1-12d=-1,(a1-d)a1-12d=0.d>0,a1=1,d=2,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(mǎn)足題設(shè),且數(shù)列bn的通項(xiàng)公式bn=4,n=1,2n-2,n2,T1=4,當(dāng)n2時(shí),Tn=4+1-2n-11-2=2n-1+3,當(dāng)n=1時(shí)也適合,數(shù)列bn的前n項(xiàng)和為T(mén)n=2n-1+3.