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经典word整理文档,仅参考,双击此处可删除页眉页脚。本资料属于网络整理,如有侵权,请联系删除,谢谢!第11123,354759Aann2n+ann2n1Cann2n-ann2n32nSnn22n1Aan2nBan2nCan2n12nDan2n12n3a11时,a1a2…an=n2,则a3a54)nn图A3B4C.5D65an111an,a82,则a16.已知数列满足:a4n31a4n10a2n=,a2009________a2014_____)ann2kn2数k的取值范围为________.8.年广东江门一模)将集合{2ss,ijb43图9.已知在等差数列和等比数列中,a1b11b48,S10(1)求和bn;(2)310an=(nn为多大时,第21)nSna12,S312a6A8BC12D2)设nS84a3,a72a9A6B4C2D23)a1,公差为-1nS1S2a1A2B2C.12D4na1a7①a21;②a7;③S13;④S14;⑤S8-其结果为确定常数的是AC5nSn12Sm0Sm13mA3B4C5D66)dAd0BCa1d0D7)a11a3=a224an_____)a3+a8103a5a7_____)在等差数列中,a38为和n10年新课标Ⅱ)已知等差数列的公差不为零,25a1,a11(1)(2)求a1a4a7a3n第31.在等比数列中,a23a7a1036a15=A12BC6D62)x,3x3,6x6A24B0C12D3的等差数列中,a1a3a1a3+a5a2a64)Aa1a3,Ba2a3,Ca2a4,Da3a6,5n8a2a50S5S2A11B5C8D6123nSnASn2an-Sn3an2CSn43anDSn37)已知是等差数列,a11na1a2S8=_____)22n(n∈N*)等于__________.9)在等比数列中,a2a12为和n10)a13,a4=12b14b420bn(1)(2)n第41a133项21a3+a4a5A33B72C84D2nSnS3a210a19a1A.13B13C.19D32a2a4,nSnAn(n1)Bn(nC.nn12D.nn4)在等比数列中,a42a558项和等于A6B5C.4D35nSna55S5=1anan+16an(1)n(3n2)a1a10=A15BC12D7)a108a13n取最大值时,nA20B21C22D8X5­4­1n为n2个数是______________.1图9年新课标Ⅰ)已知是递增的等差数列,a2x25x+60(1)(2)n10)a18b1=16anbnan1成等差数列,bn1bn1成等比数列,n1,2,3(1)求a2(2)an(3)n1a1-11a2111an-第51.在等比数列中,a11am=a1a2a3a4a5mA9BC11D21,3,6,101,4,9,16X5­5­11图①13=310;②25=916152131;⑤64=AC33bnan+1-b32b1012a8A0B3C8D4nSn8a2a50式子中数值不能确定的是C.an1nSn=23an,an=6a11an1an3an1an=7.已知在数列中,a11an2annan=8.已知在数列中,a11an3an3nan=_____a11an1=3an(1)an(2)1a11a21an10)a11nan1+n(n1)(1)(2)设bn3nann第61.观察(x2)′=2x,(x4)′=4x3(cosx)′=-Rf(x)f(x)=f(x)g(x)为f(x)g(x)Af(x)Bf(x)Cg(x)D2)n___________________________________)12112223122232=6122232-n____.4X5­6­1X5­6­1(1)c2b2.X5­6­1(2)O­ABCs1s2表示三个侧面面积,__________________.图5cosπ312cosπ5cos2π514,cosπ7cos2π7cos3π718_.6)观察下列一组等式:212=4,21×2=439292434163,43×4=163n______.7.年福建龙岩模拟)1+11+11表示无限重复)是一个固定值,可以令原式=t1+1ttt51222285④sin2(-18°)+cos248°-sin(⑤sin2(-25°)+cos255°-sin((1)5(2)(1)9)在等差数列中,a1a25,a371anan1n(1)(2)mn1mnS1Sm,Snmn第71都成立的是Alg(x2+1)≥lg(2x)B.x21C.1x21≤1D.2a2+b21-a2b2≤0,只要证A2ab1Ba2b2-1C.ab21D(a21)(b23ab,c是不全相等的实数,求证:a2b2+c2≥ab+bc+∵a,b,c∈R,∴a2+b2≥2ab,b2+c2≥2bc,a2+又abc不全相等,∴2(a2+b2+c2)≥2(ab++ac).∴a2+b2+c2≥ab+bcac.ABCD4711157111-5,7511+1,(75)2(111)2,35113511∴7-1115.ABCD5)abx2ax+b0Ax2+axb0Bx2+axb0Cx2+axb0Dx2+axb06αβ是两个不同的平面,mnα及βαβ;③n⊥β;④m⊥α.____________________.7lgx2aba+c33a3c4a2b3abc1请将错误的一个改正为________________.8){abc}{0,1,2}列三个关系:①a≠2;②b=2100a+10bc_______)|a2a3|=a1a2a3(1)(2)m1a11a2m10)d0n项Sna1=1S2S3(1)求d及Sn;(2)求mk(m,k∈N*)amam1am2amk第81.用数学归纳法证明:(n1)(n+2)…(n+n)=2n×1×3×…×(2n+1)(n∈N*),从“n=k”到“n=k+1”左端需乘的代数式是A2k1B.C.2k1D.2k3k12.用数学归纳法证明:1222n22212=n2n2+13到k时,左边应加Ak2B(k+Ck2(k+k2D(k3n与的大小关系为ABCn1或n22nnD4f(n)和g(n)①f(1)=g(1)n∈N*,f(n)f(n1)g(n)-g(n1)n∈N*时,有ABCf(n)Df(n)与g(n)51222+…+25n1是n1A12B1C122223D12226Sk=1k11k21k312k(k=1,2,3Sk1ASk12k+Sk12k21CSk12k+112k2DSk12k112k271n11n21n3m的最大值为__________.8f(n)1n1n11n1n2法正确的是________.nn2时,f(2)13;n1n=2时,f(2)1214;n2nn2f(2)12+;n2n1n2f(2)129)nSn4(n1)(Sn1)(n(1)求a1(2)求an;(3)设bnn+1annTn,求证:10)设a11an1a2n2an2+(1)若b1a2(2)若b1ca2nca2n1对第114Bk1S=2×0+11k2;S=2×1+13k3;S=2×3+17k4;S=2×7+115kS54nan11an1a82,∴a7=11a8a611a71a511a6a412a31a22a16a2009=a4×503-1a2014a1007=a4×252-17(3an1-an=(n+1)2+k(n+1)+2-n2-kn-2=2n+1+即k(2n1)k[(2n1)]max83=202152022,622;23,102123,1222232024,18=+24,2022+24b439(1)dq,则S1010a145d55⇒d1⇒an=1)dn,b4b1q38⇒q2⇒bnb1qn11,则annbn2n(2)a11a2a33b1b22b3,3(1,1)(1,2)(1,4)(2,1)(2,2)(2,4)(3,1),(3,2)(3,4)9符合题意的有(1,1)(2,2)210.解:∵an+1an(n2)1011n1(n1011n9n111011n0,an+1an0an1an;当n9an+1an0a10a9;当an1an0ann9或n=或第21Cda12S3=(a1+a3)a23a2a24da6a15d2AS8=8a1+8×72d=4a34(a12d),4a1=-20d,a15d.a1+6d=d210,a9a18d=+8×(-2)3DS1,S2成等比数列,∴S22=S1S4(2a11)2=a1(4a16)a14Aa1a7是确定的常数,故②正确;S1313a1+a132是确定的常数,故③正确;S8a6a7a85.CSm-Sm-1=am=2,Sm+1-Sm=am+1=3,am1amdSmma1mm120Sm1=m+1a1+mm+12=m6C221,21.又anan-d,故2172n1a3a22412d(1+d)24,即d24.d2an8a3a8a12da17d2a1=10,3a5a7=3(a14d)+6d)4a12(2a19d)9dn2a12d8(a13d)2(a1d)(a1+8d).a1d=4,d(d3a1)a14,d=0a11d401nSn或Sn3n210(1)da1a11成等比数列,∴a211=(a110d)2a1(a112d)d(2a125d)∵d≠0,∴2×25+25d0.d∴an=25(n2)(2)(1)a3n22(3n276n+31.可为首项,-6∴Sn=a1a4a3n2na1+a3n-n256n3n2第31Aa2a15a7a1036,则a1512.2A解析:方法一:6x63x2q3x+3x=2,3x32xx336,-12(3x+3)2=x(6x+6),9x2+18x+9=6x2+6x,3x212x9x3或x=-1()36123C4D解析:因为数列是等比数列,a26a3a9a3,a65Dq8a2a58a2a2q30,q=-2.∴S5S2=1q51q26DSna1-anq1q123an1-2337a1a2a22a1a5(a1d)2a1(a14d)d2a12S88a185686a1=2q2Sn21-2n12n12S562S669a2-a1a1(q1)∵2×2a2=3a1a3,∴4a1q=∴a1(4q-3-0q24q3q3或q=a1(q1)2q1q3a1nSn3n10.解:(1)da4a1312-33∴an=3(n设q是等比数列{bnan}有q3b4a4b1a12012438∴bn-an(b1-a1)×2n-1bnan+2n13n2n(2)nSn(36+93n)(1222n3nn12第41C2CS3=a210a1a1a3a39.a5a1q481a19a13Aa24a2a8(a16)2(a1+2)(a114),=2.∴Sn=na1nn-12dn(n1)n24Cqa5a4=52.∴a1=a4q3=∴lgan-lganlganan18lg161258(4lg23lg5)+28(lg5lg2)4lg24lg54.5Aa55S515a11d=1.∴an=1(n1)n.故1anan11nn1=1n1n+1.∴1a1a2+…+1a100a10111121100-110111101100101.6A7B5a88a135(a17d)8(a1+12d).361a1.由ana1(n1)da11)-361a1≥0⇒n≤643=故取最大值时,n21.8.n2n2a3a2=2a4a33a5a44an1n1ana223(n1)2+n-12×(n-2)n1n-22,2n22n9(1)x25x602,3,a22a4da4a22d1,∴d=12a1an32(nn2(2)nSn,(1)an2nn22n1Sn322423+n22n1,12Sn323+525n22n2,12Sn32212312412n+n+22n234+12n1n22n21n42nnSn=2n42n10(1)由2b1a1a2a22b1由a22b1b2b2a22b1(2)∵an,bn1成等差数列,∴2bn=anan+∵bn,an11成等比数列,∴a2n+1bnbn∵数列,的每一项都是正数,∴an+1bnbn时,anbn2bnbn1bn1,42∴bn=b1(n1)d2n2bn4(nan=bn-1bn=4n24n+12=4n(n+1).当n1a1=8nan4n(n1).(3)(2)1712314714n24n14n24n1271n1n∵14n2+4n-1271n1n1⇔14n24n127n2+7n⇔7n27n8n28n2⇔n2n-20⇔(n1)(n2)0时,1712314n24n11727121n1n1172712117+27×12=当n1n1a111a211a3-11an-14n24n114n24n12n-12n31412n1当17+12314n24n1171231415191711112n312n12n112n3171231415当n11727n2时,1712317=n1a111a211a3-11an-14n24n114n2112n-12n11212n1当17+12314n24n1171231471217191911112n312n12n112n117123147当n11727n2时,1712317=;当n317+12314717114n1a111a211a3-11an-第513Bb12d=-2b19d12.b16d=∵bn=an1,∴b1+b2bnan1∴a8=b1+b2+…+b73=7×(-6)+7×62×2+34D8a2a5a2(8q3)=0.∵a2≠0,∴q=-2.∴a5a3=4S5S31q51-q3=113;an+1an=q2;Sn+1Sn=1-qn+11-qn,n5(2)n-1Sn23anSn123an1+13.an=23an-23an-1,13an23an-1,anan-12,a11,则an(2)n6.13n2an1an3an11an1=3an+1an1an3⇒1an13⇒1an1+3(n1)13n7.3×2n-11an1A(n1)AnB)A1B=1.∴an+1+(n1)1=n1).∴an+n1=3×2n-1-n8n3n1解析:∵an+13an3n,∴an+13n11.令an3n1bn11的等差数列,∴bn11(n1)=n.∴an=n3n-9(1)由an13an1an112=3an+又a11232an323an12an3n-(2)(1)1an23n时,3n-1≥2×3n-,13n-1≤12×3n-11an23n-1≤13n-1a11a2+…+1an≤1+1313n13211a11a210(1)1nan11)ann(n+,n(n1)an1n1ann,即an1n1ann11(2)(1)ann1(n-1)×1=nanbn3nanSn=1×31+2×32+3×33+…+n3n,3Sn=1×32+2×33+3×34+…+n3n1,2Sn31323nn3n+1313n13n3n11-2n3n∴Sn=2n13n1第61D2.2n×1×3×5×…×(2n-1)(n3)×…×(n+31222+32(1)n1n2(1)n+1nn4s24s21s225cosπ2n+1cos2π2n+1…cosnπ2n112n6.n1n(nn+1n×(n+n1n(n1)nn2+nn=n12n,n+1n×(n+1)n12nn1n+(n1)n+1n×(n+1)72解析:类似令原式=t2tt,2tt2得t1()或t8(1)sin15°cos15°=1-12sin30°=34(2)sin2αα)sinαcos(30°-α)=sin2α+cos2(30°-α)sinαcos(30°-α)sin2α+(cos30°cosα+sin30°sinα)2-sinα(cos30°cosα+sin30°sinα)sin2α34cos2α32sinαα14sin2α-32sinαcosα12sin2α34sin2α34cos2α=9(1)d,a1a2=5a372a1d5a1=7.解得a11d=ana1+(n1)d13(n1)3nan3n(2)1anan113n-23n+1=1313n-2-13n+1,1anan1nSn1a1a2+1a2a31a3a41an1an113114+1713171313n-21313n-213n113113n+1n3nmn1mnS1Sm则S2mS1Snm3m12=14×n3n+n4m23m26mn03m26mm11m1mn4m23m26m1mnm2,n第71CAD中x0AD,B为x2B25A6(1)m⊥n,αβ,n⊥β⇒m⊥ααβ,m⊥α⇒n⊥β;(3)m⊥n,n⊥β,m⊥α⇒αβ(4)α⊥β,n⊥β,m⊥α(3)(4)(1),(2)为7lg153aclg32ablg92lg32(2ab)2blg3=2ablg94a盾.反过来,lg94alg3blg5aclg83(1lg5)3(1ac)lg5aclg83-3a3clg833a-lg5ac∴lg15=lg(3×5)=lg3lg5(2ab)(ac)3a-blg153ab8a0或a1,即a0b=1c2或a0,2c1或a=1b=0c2或a1b2c0若b2正a2b0c1100a10bc9.解:(1)∵a1a2a3=a32125又a2|q1|=1或q即q1a1=-5或q3a1∴an=-5(-1)n1或an=5×3n-(2)若q11a11a21am15m0m.若q31a11a21am1531132910110(1)S2S3(2a13d)36,将a11d2或dd0,∴d=2an12(n1)2n1,Sn12n1n2(2)(1)am1amamk[2m12mk1]k+12=k1)(k1)∵m,k∈N*,2m+k11k11,∴2m+k1=5k113.m3k12,(舍).或2mk1=13k15.m5k综上所述,m5k第815Dn125124,选6CSk+11k111k1212k11k21k12k21k11k212k112k21k1Sk12k112k7f(n)1n1n21n12n,则f(n1)-f(n)12n121n112n1-12n20{f(n)}f(n)min=f(1)=1289(1)n14×(1+1)(a1(1+2)2a1,a1当n24×(2+1)(a1a21)(22)2a2,a2(2)4(Sn1)n+22ann4(Sn11)n12an4ann22ann-n+12an1n,即anan1=n∴ann+131n3an-2n-13a233∴an=(na18a227an(nn1a18(11)3n=kak(knk+1有4(k11)(Sk11)(k2)2ak1,即4(Sk1+k2᠄

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