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1、課時(shí)知能訓(xùn)練一、選擇題1數(shù)列an中,an1,已知該數(shù)列既是等差數(shù)列又是等比數(shù)列,則該數(shù)列的前20項(xiàng)的和等于()A100B0或100C100或100 D0或1002數(shù)列an的通項(xiàng)公式an(nN*),若前n項(xiàng)和為Sn,則Sn為()A.1B.1C.(1)D.(1)3(2012惠州模擬)已知Sn為等差數(shù)列an的前n項(xiàng)和,若a12010,6,則S2011()A2011B2010C0D24已知數(shù)列an:,那么數(shù)列bn的前n項(xiàng)和Sn為()A. B.C. D.5設(shè)數(shù)列xn滿足logaxn11logaxn(nN*,a0且a1),且x1x2x3x100100,則x101x102x103x200的值為()A100a
2、2 B101a2 C100a100 D101a100二、填空題6數(shù)列3,33,333,的前n項(xiàng)和Sn_.7數(shù)列an的前n項(xiàng)和為Sn,a11,a22,an2an1(1)n(nN*),則S100_.8已知an是公差為2的等差數(shù)列,a112,則|a1|a2|a3|a20|_.三、解答題9(2012韶關(guān)模擬)已知數(shù)列an是各項(xiàng)均不為0的等差數(shù)列,Sn為其前n項(xiàng)和,且滿足aS2n1,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)數(shù)列bn滿足bn,求數(shù)列bn的前n項(xiàng)和Tn.10設(shè)函數(shù)yf(x)的定義域?yàn)镽,其圖象關(guān)于點(diǎn)(,)成中心對稱,令akf()(n是常數(shù)且n2,nN*),k1,2,n1,求數(shù)列ak的前n1
3、項(xiàng)的和11(2012汕頭模擬)已知等差數(shù)列an的前3項(xiàng)和為6,前8項(xiàng)和為4.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn(4an)qn1(q0,nN*),求數(shù)列bn的前n項(xiàng)和Sn.答案及解析1【解析】由題意知an1an0,由an1得a5an0,an5,S20100.【答案】A2【解析】an(),Sn(1)(1)(1)【答案】D3【解析】設(shè)等差數(shù)列的公差為d,則Snna1d,n2010,數(shù)列是以2010為首項(xiàng),以為公差的等差數(shù)列,由6得66,d2.S20112011(2010)20.【答案】C4【解析】an,bn4(),Sn4(1)()()4(1).【答案】B5【解析】logaxn11logaxn,
4、得xn1axn且a0,a1,xn0,數(shù)列xn是公比為a的等比數(shù)列,x101x102x103x200x1a100x2a100x3a100x100a100100a100.【答案】C6【解析】數(shù)列3,33,333,的通項(xiàng)公式an(10n1),Sn(101)(1021)(10n1)(1010210310n)n10n1.【答案】10n17【解析】由an2an1(1)n知a2k2a2k2,a2k1a2k10,a1a3a5a2n11,數(shù)列a2k是等差數(shù)列,a2k2k.S100(a1a3a5a99)(a2a4a6a100)50(246100)502 600.【答案】2 6008【解析】由題意知,an12(n1
5、)(2)2n14,令2n140,得n7,當(dāng)n7時(shí),an0;當(dāng)n7時(shí),an0.|a1|a2|a3|a20|(a1a2a7)(a8a9a20)2S7S202712(2)2012(2)224.【答案】2249【解】(1)法一設(shè)等差數(shù)列an的公差為d,首項(xiàng)為a1,在aS2n1中,令n1,n2,得即解得a11,d2,an2n1.法二an是等差數(shù)列,則a1a2n12an.S2n1(2n1)(2n1)an.由aS2n1,得a(2n1)an,又an0,an2n1,則a11,d2.an2n1.(2)bn(),Tn(1).10【解】yf(x)的圖象關(guān)于點(diǎn)(,)成中心對稱,所以f(x)f(1x)1.令Sn1a1a2an1則Sn1f()f()f(),又Sn1f()f()f(),兩式相加,得2Sn1f()f()f()f()f()f()n1,Sn1.11【解】(1)設(shè)an的公差為d.由已知得解得a13,d1.故an3(n1)4n.(2)由(1)可得,bnnqn1,于是Sn1q02q13q2nqn1.若q1,將上式兩邊同乘以q,qSn1q12q2(n1)qn1nqn.兩式相減得到(q1)Snnqn1q1q2qn1nqn于是,Sn.若q1,則Sn123n,所以,Sn