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1、新編高考數(shù)學(xué)復(fù)習(xí)資料第四節(jié)數(shù) 列 求 和全盤鞏固1(2014慈溪模擬)設(shè)等差數(shù)列an的前n項(xiàng)和是Sn,若ama10,且Sm10 BSm0CSm0,且Sm10 DSm0,且Sm10解析:選Aama10,a1am10,且Sm10.2已知an是首項(xiàng)為1的等比數(shù)列,Sn是an的前n項(xiàng)和,且9S3S6,則數(shù)列的前5項(xiàng)和為()A.或5 B.或5 C. D.解析:選C設(shè)an的公比為q,顯然q1,由題意得,所以1q39,得q2,所以是首項(xiàng)為1,公比為的等比數(shù)列,前5項(xiàng)和為.3數(shù)列12n1的前n項(xiàng)和為()A12n B22nCn2n1 Dn22n解析:選C由題意得an12n1,所以Snnn2n1.4若數(shù)列an為等
2、比數(shù)列,且a11,q2,則Tn的結(jié)果可化為()A1 B1C. D.解析:選Can2n1,設(shè)bn2n1,則Tnb1b2b3bn32n1.5已知數(shù)列an的通項(xiàng)公式為ann2cos n(nN*),Sn為它的前n項(xiàng)和,則等于()A1 005 B1 006 C2 011 D2 012解析:選B注意到cos n(1)n(nN*),故an(1)nn2.因此有S2 012(1222)(3242)(2 01122 0122)1232 0112 0121 0062 013,所以1 006.6已知數(shù)列an滿足a11,an1an2n(nN*),設(shè)Sn是數(shù)列an的前n項(xiàng)和,則S2 014()A22 0141 B321
3、0073C321 0071 D321 0072解析:選B由2,且a22,得數(shù)列an的奇數(shù)項(xiàng)構(gòu)成以1為首項(xiàng),2為公比的等比數(shù)列,偶數(shù)項(xiàng)構(gòu)成以2為首項(xiàng),2為公比的等比數(shù)列,故S2 014(a1a3a5a2 013)(a2a4a6a2 014)321 0073.7在等比數(shù)列an中,若a1,a44,則公比q_;|a1|a2|an|_.解析:設(shè)等比數(shù)列an的公比為q,則a4a1q3,代入數(shù)據(jù)解得q38,所以q2;等比數(shù)列|an|的公比為|q|2,則|an|2n1,所以|a1|a2|a3|an|(12222n1)(2n1)2n1.答案:22n18(2014衢州模擬)對(duì)于數(shù)列an,定義數(shù)列an1an為數(shù)列a
4、n的“差數(shù)列”,若a12,an的“差數(shù)列”的通項(xiàng)公式為2n,則數(shù)列an的前n項(xiàng)和Sn_.解析:an1an2n,an(anan1)(an1an2)(a2a1)a12n12n2222222n222n.Sn2n12.答案:2n12來源:數(shù)理化網(wǎng)9數(shù)列an的前n項(xiàng)和為Sn,a11,a22,an2an1(1)n(nN*),則S100_.解析:由an2an1(1)n,知a2k2a2k2,a2k1a2k10,a1a3a5a2n11,數(shù)列a2k是等差數(shù)列,a2k2k.S100(a1a3a5a99)(a2a4a6a100)來源:50(246100)502 600.答案:2 60010(2014杭州模擬)已知數(shù)列
5、an滿足a11,an11,其中nN*.(1)設(shè)bn,求證:數(shù)列bn是等差數(shù)列,并求出an的通項(xiàng)公式;(2)設(shè)cn,數(shù)列cncn2的前n項(xiàng)和為Tn,是否存在正整數(shù)m,使得Tn對(duì)于nN*恒成立?若存在,求出m的最小值;若不存在,請(qǐng)說明理由解: (1)證明:bn1bn2(常數(shù)),數(shù)列bn是等差數(shù)列a11,b12,因此bn2(n1)22n,由bn得an.(2)cn,cncn22,Tn23,依題意要使Tn0,則a9b4q2,即4q216,q2,又123945,故a50是數(shù)陣中第10行的第5個(gè)數(shù),a50b10q41024160.(2)Sn12n,Tn22.12已知數(shù)列an的前n項(xiàng)和為Snn2.設(shè)數(shù)列bn的
6、前n項(xiàng)和為Tn,且Tn(為常數(shù))令cnb2n(nN*),求數(shù)列cn的前n項(xiàng)和Rn.解:當(dāng)n1時(shí),a1S11.當(dāng)n2時(shí),anSnSn1n2(n1)22n1.當(dāng)n1時(shí),a11滿足上式an2n1(nN*)故Tn,所以n2時(shí),bnTnTn1.故cnb2n(n1)n1,nN*,所以Rn00112233(n1)n1,則Rn011223(n2)n1(n1)n,兩式相減,得Rn123n1(n1)n(n1)nn,來源:整理,得Rn.所以數(shù)列cn的前n項(xiàng)和Rn.沖擊名校1數(shù)列an滿足an1(1)nan2n1,則an的前60項(xiàng)和為()A3 690 B3 660 C1 845 D1 830解析:選D當(dāng)n2k時(shí),a2k
7、1a2k4k1,當(dāng)n2k1時(shí),a2ka2k14k3,a2k1a2k12,a2k1a2k32,a2k1a2k3,a1a5a61.a1a2a3a60(a2a3)(a4a5)(a60a61)3711(2601)30611 830.2設(shè)an是首項(xiàng)為a,公差為d的等差數(shù)列(d0),Sn是其前n項(xiàng)的和記bn,nN*,其中c為實(shí)數(shù)(1)若c0,且b1,b2,b4成等比數(shù)列,證明:Snkn2Sk(k,nN*);(2)若bn是等差數(shù)列,證明:c0.證明:由題設(shè),Snnad.(1)由c0,得bnad.又b1,b2,b4成等比數(shù)列,所以bb1b4,即2a,化簡得d22ad0.因?yàn)閐0,所以d2a.因此,對(duì)于所有的m
8、N*,有Smm2a.從而對(duì)于所有的k,nN*,有Snk(nk)2an2k2an2Sk.(2)設(shè)數(shù)列bn的公差是d1,則bnb1(n1)d1,即b1(n1)d1,nN*,代入Sn的表達(dá)式,整理得,對(duì)于所有的nN*,有n3n2cd1nc(d1b1)令A(yù)d1d,Bb1d1ad,Dc(d1b1),則對(duì)于所有的nN*,有An3Bn2cd1nD.(*)在(*)式中分別取n1,2,3,4,得ABcd18A4B2cd127A9B3cd164A16B4cd1,從而有由,得A0,cd15B,代入方程,得B0,從而cd10.即d1d0,b1d1ad0,cd10.若d10,則由d1d0,得d0,與題設(shè)矛盾,所以d10
9、.又cd10,所以c0.高頻滾動(dòng)1已知an為等比數(shù)列,Sn是它的前n項(xiàng)和若a2a32a1,且a4與2a7的等差中項(xiàng)為,則S5()A35 B33 C31 D29解析:選C設(shè)數(shù)列an的公比為q,則由等比數(shù)列的性質(zhì)知,a2a3a1a42a1,即a42.由a4與2a7的等差中項(xiàng)為知,a42a72,a7.q3,即q.a4a1q3a12,a116,S531.2已知數(shù)列an是等比數(shù)列,若a22,a5,則a1a2a2a3anan1()A16(14n) B16(12n)C.(14n) D.(12n)來源:解析:選C設(shè)等比數(shù)列an的公比為q.a22,a5,q3,a14,q,an4n1n3,anan12n58n1,a1a2a2a3anan1(14n).