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1、(福建專用)2013年高考數(shù)學(xué)總復(fù)習(xí) 第五章第4課時 數(shù)列求和課時闖關(guān)(含解析)一、選擇題1等差數(shù)列an的通項(xiàng)公式是an12n,其前n項(xiàng)和為Sn,則數(shù)列的前11項(xiàng)和為()A45B50C55 D66解析:選D.Sn,n,的前11項(xiàng)的和為66.2數(shù)列1,3,5,7,(2n1),的前n項(xiàng)和Sn為()An21 B2n2n1Cn21 Dn2n1解析:選A.該數(shù)列的通項(xiàng)公式為an(2n1),則Sn135(2n1)()n21.故選A.3(2012深圳調(diào)研)已知函數(shù)f(x)x2bx的圖象在點(diǎn)A(1,f(1)處的切線的斜率為3,數(shù)列的前n項(xiàng)和為Sn,則S2011的值為()A. B.C. D.解析:選D.f(x)
2、2xb,f(1)2b3,b1,f(x)x2x,S201111.4在數(shù)列an中,a11,a22,an2an1(1)n,那么S100的值等于()A2500 B2600C2700 D2800解析:選B.據(jù)已知當(dāng)n為奇數(shù)時, an2an0an1,當(dāng)n為偶數(shù)時,an2an2ann,故an,故S1001124610 50502600.5設(shè)數(shù)列an是首項(xiàng)為1,公比為3的等比數(shù)列,把a(bǔ)n中的每一項(xiàng)都減去2后,得到一個新數(shù)列bn,bn的前n項(xiàng)和為Sn,則對任意的nN*,下列結(jié)論正確的是()Abn13bn2,且Sn(3n1)Bbn13bn2,且Sn(3n1)Cbn13bn4,且Sn(3n1)2nDbn13bn4,
3、且Sn(3n1)2n解析:選C.因?yàn)閿?shù)列an是首項(xiàng)為1,公比為3的等比數(shù)列,所以數(shù)列an的通項(xiàng)公式為an3n1,則依題意得,數(shù)列bn的通項(xiàng)公式為bn3n12,bn13n2,3bn3(3n12)3n6,bn13bn4.bn的前n項(xiàng)和為:Sn(12)(312)(322)(332)(3n12)(13132333n1)2n2n(3n1)2n.二、選擇題6(2012三明質(zhì)檢)數(shù)列1,的前n項(xiàng)和Sn_.解析:由于an2(),Sn2(1)2(1).答案:7已知函數(shù)f(x)對任意xR,都有f(x)1f(1x),則f(2)f(1)f(0)f(1)f(2)f(3)_.解析:由條件可知:f(x)f(1x)1.而x(
4、1x)1,f(2)f(3)1,f(1)f(2)1,f(0)f(1)1,f(2)f(1)f(2)f(3)3.答案:38若數(shù)列an是正項(xiàng)數(shù)列,且n23n(nN*),則_.解析:令n1,得4,a116.當(dāng)n2時,(n1)23(n1),與已知式相減,得(n23n)(n1)23(n1)2n2,an4(n1)2,n1時,a1也適合an.an4(n1)2,4n4,2n26n.答案:2n26n三、解答題9(2011高考課標(biāo)全國卷)已知等比數(shù)列an的各項(xiàng)均為正數(shù),且2a13a21,a9a2a6.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bnlog3a1log3a2log3an,求數(shù)列的前n項(xiàng)和解:(1)設(shè)數(shù)列an的公
5、比為q,由a9a2a6得a9a,所以q2.由條件可知q0,故q.由2a13a21得2a13a1q1,所以a1.故數(shù)列an的通項(xiàng)公式為an.(2)bnlog3a1log3a2log3an(12n).2 ,故 2,所以數(shù)列的前n項(xiàng)和為.10數(shù)列an中a13,已知點(diǎn)(an,an1)在直線yx2上,(1)求數(shù)列an的通項(xiàng)公式;(2)若bnan3n,求數(shù)列bn的前n項(xiàng)和Tn.解:(1)點(diǎn)(an,an1)在直線yx2上,an1an2,即an1an2.數(shù)列an是以3為首項(xiàng),2為公差的等差數(shù)列,an32(n1)2n1.(2)bnan3n,bn(2n1)3n,Tn33532(2n1)3n1(2n1)3n,3Tn
6、332533(2n1)3n(2n1)3n1,由得2Tn332(32333n)(2n1)3n192(2n1)3n1.Tnn3n1.一、選擇題1(2012南平調(diào)研)已知數(shù)列an的通項(xiàng)公式是an,其前n項(xiàng)和Sn,則項(xiàng)數(shù)n等于()A13 B10C9 D6解析:選D.an1,Snnn1n,即n1n,解得n6.2定義:在數(shù)列an中,an0且an1,若anan1為定值,則稱數(shù)列an為“等冪數(shù)列”已知數(shù)列an為“等冪數(shù)列”,且a12,a24,Sn為數(shù)列an的前n項(xiàng)和,則S2013()A6038 B6036C2 D4解析:選A.a1a2248a2a34a3,得a32,同理得a44,a52,這是一個周期數(shù)列S20
7、13(24)26038.二、填空題3等差數(shù)列an的前n項(xiàng)和為Sn,且a4a28,a3a526.記Tn,如果存在正整數(shù)M,使得對一切正整數(shù)n,TnM都成立,則M的最小值是_解析:an為等差數(shù)列,由a4a28,a3a526,可解得a11,d4,從而Sn2n2n,Tn2,若TnM對一切正整數(shù)n恒成立,則只需Tn的最大值M即可又Tn22,只需2M,故M的最小值是2.答案:24(2012泉州質(zhì)檢)已知數(shù)列an,若a1,a2a1,a3a2,a4a3,anan1是首項(xiàng)為1,公比為2的等比數(shù)列,則an的前n項(xiàng)和Sn_.解析:由題意anan12n1,an(anan1)(an1an2)(a2a1)a12n12n2
8、2112n1,Sn(211)(221)(2n1)(21222n)n2n1n2.答案:2n1n2三、解答題5數(shù)列an滿足a11,a22,an2an2sin2,n1,2,3(1)求a3,a4及數(shù)列an的通項(xiàng)公式;(2)設(shè)Sna1a2an,求S2n.解:(1)a3a12sin2a12123,a4a22sin2a22.a2n1a2n12sin2a2n12,即a2n1a2n12,即數(shù)列a2n1是以a11,公差為2的等差數(shù)列a2n12n1.又a2n2a2n2sin2a2n,即數(shù)列a2n是首項(xiàng)為a22,公比為的等比數(shù)列,a2na2n12n1.綜上可得an.(2)S2na1a2a2n1a2na1a3a2n1a2a4a2n13(2n1)n266n.6已知數(shù)列an的每一項(xiàng)都是正數(shù),滿足a12且aanan12a0;等差數(shù)列bn的前n項(xiàng)和為Tn,b23,T525.(1)求數(shù)列an、bn的通項(xiàng)公式;(2)比較與2的大小;(3)若c恒成立,求整數(shù)c的最小值解:(1)由aanan12a0,得(an12an)(an1an)0,由于數(shù)列an的每一項(xiàng)都是正數(shù),an12an,an2n.設(shè)bnb1(n1)d,由已知有b1d3,5b1d25,解得b11,d2,bn2n1.(2)由(1)得Tnn2,當(dāng)n1時,12.當(dāng)n2時,. 122.(3)記Pn.Pn,兩式相減得Pn3.Pn遞增,Pn3,P42,最小的整數(shù)c3.