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1、高考數(shù)學(xué)精品復(fù)習(xí)資料 2019.5課時(shí)作業(yè)A組基礎(chǔ)對(duì)點(diǎn)練1數(shù)列12n1的前n項(xiàng)和為()A12nB22nCn2n1 Dn22n解析:由題意得an12n1,所以Snnn2n1.答案:C2(20xx長(zhǎng)沙模擬)已知數(shù)列an的通項(xiàng)公式是an(1)n(3n2),則a1a2a10等于()A15 B12C12 D15解析:an(1)n(3n2),a1a2a10147102528(14)(710)(2528)3515.答案:A3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()A100 B110C120 D130解析:anan1的前10項(xiàng)和為a1a2a2a3a10a
2、10a112(a1a2a10)a11a12S10102120,故選C.答案:C4已知函數(shù)yloga(x1)3(a0,a1)的圖像所過(guò)定點(diǎn)的橫、縱坐標(biāo)分別是等差數(shù)列an的第二項(xiàng)與第三項(xiàng),若bn,數(shù)列bn的前n項(xiàng)和為Tn,則T10()A. B.C1 D.解析:對(duì)數(shù)函數(shù)ylogax的圖像過(guò)定點(diǎn)(1,0),函數(shù)yloga(x1)3的圖像過(guò)定點(diǎn)(2,3),則a22,a33,故ann,bn,T1011,故選B.答案:B5.的值為 解析:設(shè)Sn,得Sn,得,Sn,Sn2.答案:26(20xx山西四校聯(lián)考)已知數(shù)列an滿足a11,an1an2n(nN*),則S2 016 .解析:數(shù)列an滿足a11,an1an
3、2n,n1時(shí),a22,n2時(shí),anan12n1,得2,數(shù)列an的奇數(shù)項(xiàng)、偶數(shù)項(xiàng)分別成等比數(shù)列,S2 016321 0083.答案:321 00837數(shù)列an滿足an1(1)nan2n1,則an的前60項(xiàng)和為 解析:當(dāng)n2k(kN*)時(shí),a2k1a2k4k1,當(dāng)n2k1(kN*)時(shí),a2ka2k14k3,a2k1a2k12,a2k3a2k12,a2k1a2k3,a1a5a61.a1a2a3a60(a2a3)(a4a5)(a60a61)3711(2601)30611 830.答案:1 8308已知數(shù)列an滿足a12a2nan(n1)2n12,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn,Tnb
4、1b2bn,求證:對(duì)任意的nN*,Tn.解析:(1)當(dāng)n1時(shí),a12a2nan(n1)2n12,a12a2(n1)an1(n2)2n2,得nan(n1)2n1(n2)2nn2n,所以an2n,n1.當(dāng)n1時(shí),a12,所以an2n,nN*.(2)證明:因?yàn)閍n2n,所以bn()因此Tn(1)()()()()(1)(),所以,對(duì)任意的nN*,Tn.9(20xx河南八市質(zhì)檢)已知遞增的等比數(shù)列an的前n項(xiàng)和為Sn,a664,且a4,a5的等差中項(xiàng)為3a3.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn,求數(shù)列bn的前n項(xiàng)和Tn.解析:(1)設(shè)等比數(shù)列an的公比為q(q0),由題意,得解得,所以an2n.(
5、2)因?yàn)閎n,所以Tn,Tn,所以Tn,故Tn.B組能力提升練1(20xx皖西七校聯(lián)考)在數(shù)列an中,an,若an的前n項(xiàng)和Sn,則n()A3 B4C5 D6解析:由an1得Snnn,則Snn,將各選項(xiàng)中的值代入驗(yàn)證得n6.答案:D2已知數(shù)列an的前n項(xiàng)和為Sn,a11,當(dāng)n2時(shí),an2Sn1n,則S2 017的值為()A2 017 B2 016C1 009 D1 007解析:因?yàn)閍n2Sn1n,n2,所以an12Snn1,n1,兩式相減得an1an1,n2.又a11,所以S2 017a1(a2a3)(a2 016a2 017)1 009,故選C.答案:C3對(duì)于數(shù)列an,定義數(shù)列an1an為數(shù)
6、列an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2 016項(xiàng)和S2 016()A22 0172 B22 0171C22 017 D22 0171解析:由題意知an1an2n,則anan12n1,an1an22n2,a3a222,a2a12,累加求和得ana12n12n22222n2,n2,又a12,所以an2n,則數(shù)列 an的前2 016項(xiàng)和S2 01622 0172.答案:A4設(shè)Sn是公差不為0的等差數(shù)列an的前n項(xiàng)和,S1,S2,S4成等比數(shù)列,且a3,則數(shù)列的前n項(xiàng)和Tn()A B.C D.解析:設(shè)an的公差為d,因?yàn)镾1a1,S22a1d2a1a1,S43
7、a3a1a1,S1,S2,S4成等比數(shù)列,所以2a1,整理得4a12a150,所以a1或a1.當(dāng)a1時(shí),公差d0不符合題意,舍去;當(dāng)a1時(shí),公差d1,所以an(n1)(1)n(2n1),所以,所以其前n項(xiàng)和Tn,故選C.答案:C5已知數(shù)列an滿足an1,且a1,則該數(shù)列的前2 016項(xiàng)的和等于 解析:因?yàn)閍1,又an1,所以a21,從而a3,a41,即得an故數(shù)列的前2 016項(xiàng)的和等于S2 0161 0081 512.答案:1 5126數(shù)列an滿足a11,nan1(n1)ann(n1),且bnancos ,記Sn為數(shù)列bn的前n項(xiàng)和,則S120 .解析:由nan1(n1)ann(n1)得1,
8、所以數(shù)列是以1為公差的等差數(shù)列,且1,所以n,即ann2,所以bnn2cos ,所以S1201222324252621202(1222232425226221202) (1222321202)3(3262921202)39(1222402)(1222321202)397 280.答案:7 2807等差數(shù)列an的前n項(xiàng)和為Sn,數(shù)列bn是等比數(shù)列,滿足a13,b11,b2S210,a52b2a3.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)若cn設(shè)數(shù)列cn的前n項(xiàng)和為Tn,求T2n.解析:(1)設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比為q.a13,b11,b2S210,a52b2a3,d2,q2.an2n1,bn2n1.(2)由(1)知,Snn(n2),cnT2n(1)(21232522n1).8已知數(shù)列an滿足n2n.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn,求數(shù)列bn的前n項(xiàng)和Sn.解析:(1)n2n,當(dāng)n2時(shí),(n1)2n1,得,2n(n2),ann2n1(n2)當(dāng)n1時(shí),11,a14也適合,ann2n1.(2)由(1)得,bnn(2)n,Sn1(2)12(2)23(2)3n(2)n,2Sn1(2)22(2)33(2)4(n1)(2)nn(2)n1,得,3Sn(2)(2)2(2)3(2)nn(2)n1n(2)n1,Sn.