《人教A版理科數(shù)學(xué)高效訓(xùn)練:55 數(shù)列的綜合應(yīng)用》由會(huì)員分享,可在線閱讀,更多相關(guān)《人教A版理科數(shù)學(xué)高效訓(xùn)練:55 數(shù)列的綜合應(yīng)用(7頁珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、 精品資料A組基礎(chǔ)演練·能力提升一、選擇題1已知數(shù)列an,bn滿足a11,且an,an1是函數(shù)f(x)x2bnx2n的兩個(gè)零點(diǎn),則b8a9()A24B32C48 D64解析:依題意有,anan1bn,an·an12n,又a11,故a22,a32,a422,a522,a623,a723,a824,a924,故b8a9(a8a9)a9a82a93×2448.答案:C2已知數(shù)列an為等差數(shù)列,數(shù)列bn是各項(xiàng)為正數(shù)的等比數(shù)列,其公比q1,若a4b4,a12b12,則()Aa8b8 Ba8>b8Ca8<b8 Da8>b8或a8<b8解析:bn為等比數(shù)
2、列,其公比q1,b4b12,a4a12,a8 > b8.答案:B3已知正項(xiàng)等差數(shù)列an滿足:an1an1a(n2),等比數(shù)列bn滿足:bn1bn12bn(n2),則log2(a2b2)()A1或2 B0或2C2 D1解析:由題意可知,an1an12ana,解得an2(n2)(由于數(shù)列an每項(xiàng)都是正數(shù)),又bn1bn1b2bn(n2),所以bn2(n2),log2(a2b2)log242.答案:C4(2013年高考遼寧卷)下面是關(guān)于公差d>0的等差數(shù)列an的四個(gè)命題:P1:數(shù)列an是遞增數(shù)列;P2:數(shù)列nan是遞增數(shù)列;P3:數(shù)列是遞增數(shù)列;P4:數(shù)列an3nd是遞增數(shù)列其中的真命題
3、為()Ap1,p2 Bp3,p4Cp2,p3 Dp1,p4解析:設(shè)ana1(n1)ddna1d,它是遞增數(shù)列,所以p1為真命題;若an3n12,則滿足已知,但nan3n212n并非遞增數(shù)列,所以p2為假命題;若ann1,則滿足已知,但1是遞減數(shù)列,所以p3為假命題;設(shè)an3nd4dna1d,它是遞增數(shù)列,所以p4為真命題答案:D來源:數(shù)理化網(wǎng)5(2014年保定調(diào)研)在ABC中,角A,B,C所對(duì)的邊分別是a,b,c,且acos C,bcos B,ccos A成等差數(shù)列,若b,則ac的最大值為()A. B3C2 D9來源:解析:acos C,bcos B,ccos A成等差數(shù)列,2bcos Bac
4、os Cccos A,cos B,b2a2c22accos B(ac)23ac(ac)23·2,即3,當(dāng)且僅當(dāng)ac時(shí),等號(hào)成立,ac2.答案:C6若關(guān)于x的方程x2xa0與x2xb0(ab)的四個(gè)根組成首項(xiàng)為的等差數(shù)列,則ab的值是()A. B.C. D.解析:設(shè)兩個(gè)方程的根分別為x1、x4和x2、x3.因?yàn)閤1x4x2x31,所以x1,x4,從而x2,x3.則ax1x4,bx2x3,或a,b,ab.答案:D二、填空題7(2013年高考重慶卷)已知an是等差數(shù)列,a11,公差d0,Sn為其前n項(xiàng)和,若a1,a2,a5成等比數(shù)列,則S8_.解析:因?yàn)閍n為等差數(shù)列,且a1,a2,a5成
5、等比數(shù)列,所以a1(a14d)(a1d)2,解得d2a12,所以S864.答案:648九章算術(shù)之后,人們進(jìn)一步用等差數(shù)列求和公式來解決更多的問題,張丘建算經(jīng)卷上第22題為:“今有女善織,日益功疾,且從第2天起,每天比前一天多織相同量的布,若第一天織5尺布,現(xiàn)在一月(按30天計(jì)),共織390尺布”,則每天比前一天多織_尺布(不作近似計(jì)算)解析:由題意知,a15,n30,Sn39030×5dd.答案:9(2014年合肥模擬)已知數(shù)列an滿足anan1an2an324,且a11,a22,a33,則a1a2a3a2 013_.來源:解析:由anan1an2an324可知,an1an2an3&
6、#183;an424,得an4an,所以數(shù)列an是周期為4的數(shù)列,再令n1,求得a44,每四個(gè)一組可得(a1a2a3a4)(a2 009a2 010a2 011a2 012)a2 01310×50315 031.答案:5 031三、解答題10(2014年大同模擬)已知公比為q的等比數(shù)列an的前6項(xiàng)和S621,且4a1,a2,a2成等差數(shù)列(1)求an;(2)設(shè)bn是首項(xiàng)為2,公差為a1的等差數(shù)列,其前n項(xiàng)和為Tn,求不等式Tnbn>0的解集解析:(1)4a1,a2,a2成等差數(shù)列,4a1a23a2,即4a12a2,q2.則S621,解得a1,an.(2)由(1)得a1,bn2(
7、n1),Tn2n(n1)·,Tnbn>0,即>0,解得1<n<14(nN*),故不等式Tnbn>0的解集為nN*|1<n<1411已知單調(diào)遞增的等比數(shù)列an滿足a2a3a428,且a32是a2,a4的等差中項(xiàng)(1)求數(shù)列an的通項(xiàng)公式;(2)若bnanlogan,Snb1b2bn,求使Snn·2n1>50成立的正整數(shù)n的最小值解析:(1)設(shè)等比數(shù)列an的首項(xiàng)為a1,公比為q,依題意,有2(a32)a2a4,代入a2a3a428,得a38,a2a420,解得或.又?jǐn)?shù)列an單調(diào)遞增,q2,a12,an2n.(2)由題意知bn2n&
8、#183;log2nn·2n,Sn1×22×223×23n×2n,2Sn1×222×233×24(n1)×2nn×2n1,得Sn222232nn·2n1n·2n12n1n·2n12,Snn·2n1>50,2n12>50,2n1>52,又當(dāng)n4時(shí),2n12532<52,當(dāng)n5時(shí),2n12664>52.故使Snn·2n1>50成立的正整數(shù)n的最小值為5.12(能力提升)(2013年高考廣東卷)設(shè)數(shù)列an的前n項(xiàng)和為
9、Sn.已知a11,an1n2n,nN*.(1)求a2的值;(2)求數(shù)列an的通項(xiàng)公式;(3)證明:對(duì)一切正整數(shù)n,有<.解析:(1)依題意,2S1a21,又S1a11,所以a24.(2)當(dāng)n2時(shí),2Snnan1n3n2n,2Sn1(n1)an(n1)3(n1)2(n1),兩式相減得2annan1(n1)an(3n23n1)(2n1),整理得(n1)annan1n(n1),即1,又1,故數(shù)列是首項(xiàng)為1,公差為1的等差數(shù)列,來源:所以1(n1)×1n,所以ann2.(3)證明:當(dāng)n1時(shí),1<;當(dāng)n2時(shí),1<;當(dāng)n3時(shí),<,此時(shí)1<11<.綜上,對(duì)一切正
10、整數(shù)n,有<.B組因材施教·備選練習(xí)1各項(xiàng)都是正數(shù)的等比數(shù)列an的公比q1,且a2,a3,a1成等差數(shù)列,則q的值為()A. B.C. D.或解析:a2,a3,a1成等差數(shù)列,a3a1a2,q21q,q或q,等比數(shù)列an的各項(xiàng)都是正數(shù),q不滿足題意,舍去,q.答案:C來源:數(shù)理化網(wǎng)2(2014年成都模擬)已知數(shù)列an滿足an2an1an1an,nN*,且a5.若函數(shù)f(x)sin 2x2cos2,記ynf(an),則數(shù)列yn的前9項(xiàng)和為()A0 B9C9 D1解析:由數(shù)列an滿足an2an1an1an,nN*可知該數(shù)列是等差數(shù)列,根據(jù)題意可知只要該數(shù)列中a5,數(shù)列yn的前9項(xiàng)和
11、就能計(jì)算得到一個(gè)定值,又因?yàn)閒(x)sin 2x1cos x,則可令數(shù)列an的公差為0,則數(shù)列yn的前9項(xiàng)和為S9(sin 2a1sin 2a2sin 2a9)(cos a1cos a2cos a9)99sin 2a59cos a599sin9cos 99.答案:C3已知數(shù)列an的前n項(xiàng)和為Sn,點(diǎn)(n,Sn)(nN*)在函數(shù)f(x)x2x的圖象上(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)數(shù)列的前n項(xiàng)和為Tn,不等式Tn>loga(1a)對(duì)任意正整數(shù)n恒成立,求實(shí)數(shù)a的取值范圍解析:(1)點(diǎn)(n,Sn)在函數(shù)f(x)x2x的圖象上,Snn2n,當(dāng)n2時(shí),Sn1(n1)2(n1),得ann.當(dāng)n1時(shí),a1S11,符合上式,ann.(2)由(1)得,Tn.Tn1Tn>0,數(shù)列Tn單調(diào)遞增(Tn)minT1.要使不等式Tn>loga(1a)對(duì)任意正整數(shù)n恒成立,只要>loga(1a)即可1a>0,0<a<1,1a>a,即0<a<,實(shí)數(shù)a的取值范圍是.