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1、 課時(shí)分層訓(xùn)練(三十三)數(shù)列求和A組基礎(chǔ)達(dá)標(biāo)一、選擇題1數(shù)列1,3,5,7,(2n1),的前n項(xiàng)和Sn的值等于()An21B2n2n1Cn21Dn2n1A該數(shù)列的通項(xiàng)公式為an(2n1),則Sn135(2n1)n21.2在數(shù)列an中,an1an2,Sn為an的前n項(xiàng)和若S1050,則數(shù)列anan1的前10項(xiàng)和為()A100B110 C120D130Canan1的前10項(xiàng)和為a1a2a2a3a10a112(a1a2a10)a11a12S10102120.故選C.3中國(guó)古代數(shù)學(xué)著作算法統(tǒng)宗中有這樣一個(gè)問(wèn)題:“三百七十八里關(guān),初行健步不為難,次日腳痛減一半,六朝才得到其關(guān),要見(jiàn)次日行里數(shù),請(qǐng)公仔細(xì)算
2、相還”其意思為:有一個(gè)人走378里路,第一天健步行走,從第二天起腳痛每天走的路程為前一天的一半,走了6天后到達(dá)目的地,請(qǐng)問(wèn)第二天走了()【導(dǎo)學(xué)號(hào):79140183】A192里B96里 C48里D24里B由題意,知每天所走路程形成以a1為首項(xiàng),公比為的等比數(shù)列,則378,解得a1192,則a296,即第二天走了96里故選B.4已知數(shù)列5,6,1,5,該數(shù)列的特點(diǎn)是從第二項(xiàng)起,每一項(xiàng)都等于它的前后兩項(xiàng)之和,則這個(gè)數(shù)列的前16項(xiàng)之和S16等于()A5B6 C7D16C根據(jù)題意這個(gè)數(shù)列的前8項(xiàng)分別為5,6,1,5,6,1,5,6,發(fā)現(xiàn)從第7項(xiàng)起,數(shù)字重復(fù)出現(xiàn),所以此數(shù)列為周期數(shù)列,且周期為6,前6項(xiàng)和
3、為561(5)(6)(1)0.又因?yàn)?6264,所以這個(gè)數(shù)列的前16項(xiàng)之和S162077.故選C.5已知函數(shù)f(x)xa的圖像過(guò)點(diǎn)(4,2),令an,nN,記數(shù)列an的前n項(xiàng)和為Sn,則S2 019()A.1B1C.1D1C由f(4)2得4a2,解得a,則f(x)x.所以an,S2 019a1a2a3a2 019()()()()1.二、填空題6設(shè)數(shù)列an 的前n項(xiàng)和為Sn,且ansin,nN,則S2 018_.1ansin,nN,顯然每連續(xù)四項(xiàng)的和為0.S2 018S4504a2 017a2 0180101.7計(jì)算:321422523(n2)2n_.4設(shè)S345(n2),則S345(n2).兩
4、式相減得S3.所以S334.8(20xx全國(guó)卷)等差數(shù)列an的前n項(xiàng)和為Sn,a33,S410,則_.設(shè)等差數(shù)列an的公差為d,則由得Snn11,2. 22.三、解答題9(20xx南京、欽州第二次適應(yīng)性考試)已知數(shù)列an的前n項(xiàng)和Sn滿足:Snn22n,nN.(1)求數(shù)列an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和. 【導(dǎo)學(xué)號(hào):79140184】解(1)當(dāng)n2時(shí),anSnSn12n1,a1S13也滿足an2n1,所以數(shù)列an的通項(xiàng)公式為an2n1.(2)由(1)知,則Tn.10(20xx太原模擬(二)已知數(shù)列an的前n項(xiàng)和Sn,數(shù)列bn滿足bnanan1(nN)(1)求數(shù)列bn的通項(xiàng)公式;(2)若c
5、n2(bn1)(nN),求數(shù)列cn的前n項(xiàng)和Tn.解(1)當(dāng)n1時(shí),a1S11,當(dāng)n2時(shí),anSnSn1n,當(dāng)n1時(shí),a11,符合上式,ann(nN),bnanan12n1.(2)由(1)得ann,bn2n1,cn2(bn1)n2n1,Tn122223324n2n1,2得2Tn123224325n2n2,得Tn22232n1n2n2(1n)2n24,Tn(n1)2n24.B組能力提升11(20xx石家莊一模)已知函數(shù)f(x)的圖像關(guān)于x1對(duì)稱,且f(x)在(1,)上單調(diào),若數(shù)列an是公差不為0的等差數(shù)列,且f(a50)f(a51),則an的前100項(xiàng)的和為()A200B100C0D50B因?yàn)楹?/p>
6、數(shù)f(x)的圖像關(guān)于x1對(duì)稱,又函數(shù)f(x)在(1,)上單調(diào),數(shù)列an是公差不為0的等差數(shù)列,且f(a50)f(a51),所以a50a512,所以S10050(a50a51)100,故選B.12(20xx合肥二次質(zhì)檢)已知數(shù)列an的前n項(xiàng)和為Sn,若Sn2an2n,則Sn_. 【導(dǎo)學(xué)號(hào):79140185】n2n(nN)由Sn2an2n得當(dāng)n1時(shí),S1a12;當(dāng)n2時(shí),Sn2(SnSn1)2n,即1,所以數(shù)列是首項(xiàng)為1,公差為1的等差數(shù)列,則n,Snn2n(n2),當(dāng)n1時(shí),也符合上式,所以Snn2n(nN)13(20xx廣州綜合測(cè)試(二)設(shè)Sn是數(shù)列an的前n項(xiàng)和,已知a13,an12Sn3(
7、nN)(1)求數(shù)列an的通項(xiàng)公式;(2)令bn(2n1)an,求數(shù)列bn的前n項(xiàng)和Tn.解(1)當(dāng)n2時(shí),由an12Sn3得an2Sn13,兩式相減,得an1an2Sn2Sn12an,an13an,3.當(dāng)n1時(shí),a13,a22S132a139,則3.數(shù)列an是以a13為首項(xiàng),公比為3的等比數(shù)列an33n13n.(2)法一:由(1)得bn(2n1)an(2n1)3n,Tn13332533(2n1)3n,3Tn132333534(2n1)3n1,得2Tn1323223323n(2n1)3n132(32333n)(2n1)3n132(2n1)3n16(2n2)3n1.Tn(n1)3n13.法二:由(1)得bn(2n1)an(2n1)3n.(2n1)3n(n1)3n1(n2)3n,Tnb1b2b3bn(03)(330)(23433)(n1)3n1(n2)3n(n1)3n13.