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1、專題升級(jí)訓(xùn)練10數(shù)列的求和及其綜合應(yīng)用(時(shí)間:60分鐘滿分:100分)一、選擇題(本大題共6小題,每小題6分,共36分)1等差數(shù)列an滿足a2a9a6,則S9()A2 B0C1 D22已知Sn為等差數(shù)列an的前n項(xiàng)和,若a12 010,6,則S2 012()A2 011 B2 010C2 012 D03已知Sn是非零數(shù)列an的前n項(xiàng)和,且Sn2an1,則S2 012()A122 012 B22 0121C22 0111 D22 0124等差數(shù)列an的前n項(xiàng)和為Sn,已知a5a74,a6a82,則當(dāng)Sn取最大值時(shí)n的值是()A5 B6C7 D85設(shè)Sn為等差數(shù)列an的前n項(xiàng)和,若a11,公差d2
2、,Sk2Sk24,則k()A8 B7C6 D56若向量an(cos 2n,sin n),bn(1,2sin n)(nN*),則數(shù)列anbn2n的前n項(xiàng)和Sn()An2 Bn22nC2n24n Dn2n二、填空題(本大題共3小題,每小題6分,共18分)7已知an是等差數(shù)列,Sn為其前n項(xiàng)和,nN*.若a316,S2020,則S10的值為_8已知數(shù)列an滿足a1,且對(duì)任意的正整數(shù)m,n都有amnaman,則數(shù)列an的前n項(xiàng)和Sn_.9對(duì)于數(shù)列an,定義數(shù)列an1an為數(shù)列an的“差數(shù)列”,若a12,an的“差數(shù)列”的通項(xiàng)為2n,則數(shù)列an的前n項(xiàng)和Sn_.三、解答題(本大題共3小題,共46分解答應(yīng)
3、寫出必要的文字說明、證明過程或演算步驟)10(本小題滿分15分)已知在數(shù)列an中,a1,an1(nN*)(1)求數(shù)列an的通項(xiàng)公式;(2)已知bn的前n項(xiàng)和為Sn,且對(duì)任意正整數(shù)N*,都有bn1成立求證:Sn1.11.(本小題滿分15分)已知數(shù)列an是公比為d(d1)的等比數(shù)列,且a1,a3,a2成等差數(shù)列(1)求d的值;(2)設(shè)數(shù)列bn是以2為首項(xiàng),d為公差的等差數(shù)列,其前n項(xiàng)和為Sn,試比較Sn與bn的大小12(本小題滿分16分)已知等差數(shù)列an的公差d0,它的前n項(xiàng)和為Sn,若S570,且a2,a7,a22成等比數(shù)列(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,求證:Tn.參
4、考答案一、選擇題1B解析:方法一:a2a9a6,a1da18da15d,即a14d.S99a136d9(4d)36d0.故選B.方法二:由a2a9a6,得a53da54da5d,a50.則S99a50,故選B.2C解析:設(shè)數(shù)列an的公差為d,則n,63d.d2.故Snna1n2nn(na11)S2 0122 012.故選C.3B解析:Sn2an1,Sn12an11(n2),兩式相減,得an2an2an1,即an2an1,數(shù)列an是公比為2的等比數(shù)列由S12a11得a11,S2 01222 0121.故選B.4B解析:由a5a74,a6a82,兩式相減,得2d6,d3.a5a74,2a64,即a
5、62.由a6a15d,得a117.ana1(n1)(3)203n.令an0,得n,前6項(xiàng)和最大,故選B.5D解析:由Sk2Sk24,ak1ak224,a1kda1(k1)d24,2a1(2k1)d24.又a11,d2,k5.6B解析:anbn2ncos 2n2sin2n2n(12sin2n)2sin2n2n2n1,則數(shù)列anbn2n是等差數(shù)列,Snn22n,故選B.二、填空題7110解析:設(shè)等差數(shù)列an的首項(xiàng)為a1,公差為d,由題意得解之得a120,d2.S101020(2)110.82解析:令m1,則an1a1an,數(shù)列an是以a1為首項(xiàng),為公比的等比數(shù)列Sn2.92n12解析:an1an2
6、n,當(dāng)n2時(shí),an(anan1)(an1an2)(a2a1)a12n12n2222222n.當(dāng)n1時(shí),a12也適合上式,an2n(nN*)Sn2n12.三、解答題10(1)解:an1(nN*),即.數(shù)列是以2為首項(xiàng),為公差的等差數(shù)列,故2.an.(2)證明:bn1,bn.Snb1b2bn1,Sn1.11解:(1)2a3a1a2,2a1d2a1a1d.2d2d10.d1,d.(2)bn2(n1),Sn.Snbn.n1或n10時(shí),Snbn;2n9時(shí),Snbn;n11時(shí),Snbn.12(1)解:因?yàn)閿?shù)列an是等差數(shù)列,所以ana1(n1)d,Snna1d.依題意,有即解得a16,d4或a114(舍去),d0(舍去)所以數(shù)列an的通項(xiàng)公式為an4n2(nN*)(2)證明:由(1)可得Sn2n24n,所以.所以Tn.因?yàn)門n0,所以Tn.因?yàn)門n1Tn0,所以數(shù)列Tn是遞增數(shù)列,所以TnT1.所以Tn.