課時作業(yè)(二十八)1.C[解析]由題意,得a2=2a1+1=3,a3=2a2+1=7,a4=2a3+1=15,故選C.2.B[解析]易知數(shù)列1,4,9,16,25,…的一個通項公式為an=-1n·n2,故選3.B[解析]觀察數(shù)列,得出規(guī)律:a2a1=21,a3a2=22,a4a3=23,a5a4=24,因此a6a5=25,所以a6=62,故選B.4.5[解析]因為an=3n228n=3n14321963,且n∈N*,所以當(dāng)n=5時,an取得最小值.5.2n[解析]當(dāng)n≥2時,an=SnSn1=n2+n(n1)2(n1)=2n.當(dāng)n=1時,a1=S1=2,滿足上式.故an=2n.6.C[解析]當(dāng)n≥2時,Sn=n+23an,Sn1=n+13an1.兩式作差可得an=SnSn1=n+23ann+13an1,則anan-1=n+1n-7.A[解析]∵(3n+2)an+1=(3n1)an,∴an+1=3n-13n+2an,∴an=3(n-1)-13(n-1)+2·3(n-2)-138.D[解析]根據(jù)題意可知a1=3,a2=6,an+2=an+1an,那么a3=a2a1=3,a4=a3a2=3,a5=a4a3=6,a6=a5a4=3,a7=a6a5=3,a8=a7a6=6,…,可知數(shù)列{an}的周期為6,那么a2018=a336×6+2=a2=6,故選D.9.D[解析]∵an=8+2n-72n,∴an+1=8+2n-52n+1,∴an+1an=2n-52n+12n-72n=2n-5-2(2n-7)2n+1=-2n+92n+1.∴當(dāng)1≤n≤4時,an+1>an,即a5>a4>a3>a2>a1;當(dāng)n≥5時,an+1<an,即a5>a6>a7>….因此數(shù)列an先遞增后遞減,∴當(dāng)n=5時,a10.D[解析]根據(jù)題意可構(gòu)造數(shù)列{an2bn},則an+12bn+1=an+2bn2an2bn=(12)an(12)2bn=(12)(an2bn).因為a1=b1=1,所以a12b1=12,所以{an2bn}是以12為首項,12為公比的等比數(shù)列,故an2bn=(12)n,所以A,B不正確.因為{an2bn}的公比為12,其絕對值小于1,所以{|an2bn|}為遞減數(shù)列,所以C不正確.anbn2=1bn·|an2bn|,易知數(shù)列an,bn為遞增數(shù)列,故1bn為遞減數(shù)列,又{|an2bn|}為遞減數(shù)列,故anbn11.an=1,n=1,2n-2,n≥2[解析]由an+1=Sn①,可得an=Sn1(n≥2)②,①②得an+1an=SnSn1=an(n≥2),即an+1an=2(n≥2),又a2=S1=1,所以a2a1=1≠12.53,4[解析]因為an是遞減數(shù)列,數(shù)列{an}從a4項開始用式子(t13)x-3計算,所以只要t13<0,即t<13即可.因為a1,a2,a3通過x23tx+18計算,所以根據(jù)二次函數(shù)的性質(zhì),應(yīng)該有3t2>52且a3>a4,即t>53且99t+18>t13,解得5313.n3n-2[解析]由anan+1=2anan+1n(n+1)可得1an+11an=2n(n+1)=21n1n+1,利用累加法可得1an1an-1+1an-11an-2+…+1a21a1=214.解:(1)設(shè)等差數(shù)列{an}的公差為d,由a3=24,S11=0,可得a1+2d=24,11a(2)由(1)知Sn=4n2+44n=4n1122+121,因為n∈N*,所以當(dāng)n=5或6時,Sn取得最大值.15.解:(1)因為a2=72,所以由a2=3a11可求得a1=3因為an+1=3an1,所以an+112=3an12,所以數(shù)列an12是以1為首項,以3為公比的等比數(shù)列.所以an12=3n1,即an=12+3n1.故Sn=n2+3(2)依題意,3n-1+13n-1≤m,即13+4設(shè)cn=13+43(3n-1),則易知數(shù)列cn綜上,可得m≥1.故所求實數(shù)m的取值范圍是[1,+∞).16.解:(1)證明:由題知bn+1+2bn+2=2bn+2+2bn+2=2,∴b1+2=4,∴數(shù)列{bn+2}是以4為首項以2為公比的等比數(shù)列.(2)由(1)可得bn+2=4·2n1,故bn=2n+12.∵an+1an=bn,∴a2a1=b1,a3a2=b2,a4a3=b3,…,anan1=bn1,累加得ana1=b1+b2+b3+…+bn1=(222)+(232)+(242)+…+(2n2)=22(1-2n-1)1-22(n1)=2n+12加練一課(四)1.A[解析]∵an+1=2an1,∴an+11=2(an1).∵a11=0,∴an1=0,即an=1,故選A.2.C[解析]a4=a2+a2=12,a6=a4+a2=18,a10=a6+a4=30.故選C.3.D[解析]由an+1=Sn+1①,可得an=Sn1+1(n≥2)②,①②得an+1=2an,又∵a2=S1+1=3,a1=2,∴S5=2+3×(1-24)4.B[解析]因為數(shù)列{an}前8項的值各異,且an+8=an對任意的n∈N*都成立,所以該數(shù)列為周期為8的周期數(shù)列.為使數(shù)列中可取遍{an}前8項的值,必須保證項數(shù)被8除的余數(shù)可以取到0,1,2,3,4,5,6,7.經(jīng)驗證A,C,D都不可以,因為它們的項數(shù)全部由奇數(shù)組成,被8除的余數(shù)只能是奇數(shù),故選B.5.B[解析]由條件知an+1an=n+1n+2,分別令n=1,2,3,…,(n1)(n≥2),可得a2a1=23,a3a2=34,a4a3=45,…,anan-1=nn+1,累乘得a2a1·a3a2·a4a6.D[解析]∵數(shù)列an滿足a1=1,an+1·an=2n(n∈N*),∴a2·a1=2,解得a2=2.由題得an+1anan+2an+1=2n2n+1,即an+2an=2,∴數(shù)列{an}的奇數(shù)項與偶數(shù)項分別成等比數(shù)列,首項分別為a1=1,a2=2,公比都為2,則S2017=(a1+a3+…+a2017)+(a27.C[解析]由條件知an+1an=1n+1+n=n+1n.分別令n=1,2,3,…,(n1),代入上式得到(n1)個等式,這些等式累加可得(a2a1)+(a3a2)+(a4a3)+…+(anan1)=(21)+(32)+(43)+…+(nn-1),即ana1=n1.又因為a1=1,所以a8.D[解析]因為anan+1=nanan+1,所以an-an+1anan+1=1an+11an=n,所以1an=1an1an-1+1an-11an-2+…+1a21a1+1a1=(9.D[解析]由已知得a3+a1=(a3+a2)(a2a1)=1,同理可得a5+a7=1,…,a37+a39=1,又a2+a4=(a3+a2)+(a4a3)=2+3=5,a6+a8=13,…,a38+a40=77,∴S40=(a1+a3+…+a39)+(a2+a4+…+a40)=10×1+(5+13+…+77)=10+410=420,故選D.10.C[解析]由an=2nan-1an-1+n-1,得nan=n-12an-1+12,于是nan1=12n-1an-11(n≥2,n∈N*).又1a11=12,∴數(shù)列nan1是以12為首項,12為公比的等比數(shù)列,故nan1=12n(11.B[解析]∵a1=12,a2=12,an+1=2an+an1,∴an+1-an-12an=1,a3=2a2+a1=32,∴1ai-1ai+1=1ai-1ai+1·ai+1-ai-12ai=12ai1ai-11ai+1=121ai-1ai1aiai+1,∑12.B[解析]根據(jù)題意,數(shù)列an滿足an=2an-1,n≥6,an-1+1,2≤n<6,且a1=a,則a2=a1+1=a+1,a3=a2+1=a+2,a4=a3+1=a+3,a5=a4+1=a+4,a6=2a5=2a+8,a7=2a6,…對于①,當(dāng)a=4時,a6=2a+8=0,此時數(shù)列an+5不是等比數(shù)列,故①錯誤;對于②,若S5<100,則有S5=(a1+a2+…+a5)=5(a+2)<100,則有a<18,故②正確;對于③,根據(jù)題意,a3=a+2,a6=2a+8,a9=24a5=16×(a+4),若a3,a6,a9成等比數(shù)列,則有(2a+8)2=(a+2)×16×(a+4),13.19[解析]因為an+1=an+2,所以an+1an=2,所以數(shù)列an是首項為1,公差為2的等差數(shù)列,所以a10=1+(101)×2=1914.22017[解析]由題得,a2=3a1+2b1=5,當(dāng)n≥2時,an+1=3an+2bn=3an2an1,所以an+1an=2(anan1),又a2a1=4,所以數(shù)列{anan1}是首項為4,公比為2的等比數(shù)列,所以a2017a2016=4×220161=22017.15.1,n=1,n!2,n≥2[解析]當(dāng)n≥2時,由已知得an+1=a1+2a2+3a3+…+(n1)an1+nan,用此等式減去已知等式,得an+1an=nan,即an+1=(n+1)an,又a2=a1=1,∴a1=1,a2a1=1,a3a2=3,a4a3=4,…,anan-1=n,將以上n個式子相乘,得an16.4017[解析]設(shè)該數(shù)列為{an},則a1=2008,a2=2009,a3=1,a4=2008,由題意得a5=2009,a6=1,a7=2008,…所以an+6=an,即數(shù)列an是以6為周期的數(shù)列,又a1+a2+a3+a4+a5+a6=0,∴S2018=336(a1+a2+a3+a4+a5+a6)+(a1+a2)=4017課時作業(yè)(二十九)1.B[解析]設(shè)等差數(shù)列{an}的公差為d,由題設(shè)可得3×(2)+3×22·d=0,解得d=2,故選B2.B[解析]根據(jù)等差數(shù)列的性質(zhì)可得,等差數(shù)列第1,4,7項的和,第2,5,8項的和與第3,6,9項的和成等差數(shù)列,所以a3+a6+a9=2×3339=27,故選B.3.C[解析]S12=(a1+a12)×122=6×(a4+a4.9[解析]根據(jù)等差數(shù)列的性質(zhì)可知a3+a4+a5+a6+a7=5a5=45,所以a5=9.5.7[解析]由(a5+a8)(a3+a6)=3911=4d=28,得d=7.6.B[解析]由a2+a4+a6=3得a4=1,則a1+a3+a5+a7=4a4=4,故選B.7.A[解析]設(shè)等差數(shù)列{an}的公差為d.可以令i=1,j=2,k=3,l=4,則aialajak=a1a4a2a3=a1(a1+3d)(a1+d)(a1+2d)=2d2≤0,S1S4S2S3=a1(4a1+6d)(2a1+d)(3a1+3d)=2a123a1d3d2=2a1+34d2158d2≤0,故只有A8.B[解析]設(shè)等差數(shù)列{an}的公差為d,由題意,得2a1+4d=-14,9a1+36d=-27,解得a1=-11,d=2,則an=2n13.令an=2n-9.A[解析]設(shè)最上面一節(jié)竹子的容積為a1,則依題意可知a1+a2+a3+a4=3,a7+a8+a9=4.根據(jù)等差數(shù)列的性質(zhì)可知a1+a2+a3+a4=2(a2+a3)=3,a7+a8+a9=3a8=4,則有a2+a3=3210.A[解析]設(shè)等差數(shù)列{bn}的公差為d,則由題設(shè)可得bn=an+1an=b1+(n1)d,則a2a1=b1,a3a2=b1+d,a4a3=b1+2d,…,a31a30=b1+29d,累加得a31a1=30b1+(1+2+…+29)d=30b1+29×302d,即a31=15(2b1+29d),又b15+b16=2b1+29d=15,所以a31=15(2b1+29d)=15×15=225,故選A11.2017[解析]∵Sn是等差數(shù)列an的前n項和,∴Snn是等差數(shù)列,設(shè)其公差為d.∵S20142014S20082008=6,∴6d=6,d=1.∵a1=2017,∴S11=2017,∴Snn=2017+(n1)×1=2018+n,∴S201712.201732016[解析]由題意可得,當(dāng)n≥2時,an+bn=an1+bn1+2,anbn=13an113bn1=13(an1bn1),所以數(shù)列{an+bn}是以a1+b1=3為首項,2為公差的等差數(shù)列,數(shù)列{anbn}是以a1b1=1為首項,13為公比的等比數(shù)列,所以(a1008+b1008)(a2017b2017)=(3+2×1007)×1×13.解:(1)設(shè)等差數(shù)列an的公差為d因為a1+a2=6,a2+a3=10,所以a3a1=4,所以2d=4,d=2.又a1+a2=a1+a1+d=6,所以a1=2,所以an=a1+(n1)d=2n.(2)記bn=an+an+1,所以bn=2n+2(n+1)=4n+2,又bn+1bn=4(n+1)+24n2=4,所以數(shù)列bn是首項為6,公差為4的等差數(shù)列其前n項和Sn=n(b1+bn)2=14.解:(1)由題意知2Sn=1+an,即4Sn=(1+an)2.當(dāng)n=1時,可得a1=1當(dāng)n≥2時,有4Sn1=(an1+1)2,又4Sn=(an+1)2,兩式相減得(an+an1)(anan12)=0,∵an>0,∴anan1=2,則數(shù)列{an}是以1為首項,2為公差的等差數(shù)列,即an=2n1.(2)2anan+1=∴Tn=113+1315+…+12n-112n+1=115.C[解析]設(shè)等差數(shù)列{an}的公差為d.∵S3=a1+a2+a3=3a2=9,a2a4=21,∴a2=3,a4=7,d=2,an=2n1.設(shè)Tn=b1a1+b2a2+…+bnan=b11+b23+…+bn2n-1=112n,則Tn+1=b11+b23+…+bn2n-1+bn+12n+1=112n+1,兩式作差得Tn+1Tn=bn+116.8[解析]由題知當(dāng)n為奇數(shù)時,bn=2,當(dāng)n為偶數(shù)時,bn=3.又a2b1=b2a1+b1,可得a2=74.當(dāng)n=2k時,有a2k+1b2k=b2k+1a2k+b2k,即3a2k+1=2a2k+3①.當(dāng)n=2k1時,有a2kb2k1=b2ka2k1+b2k1,即2a2k=3a2k12②.當(dāng)n=2k+1時,有a2k+2b2k+1=b2k+2a2k+1+b2k+1,即2a2k+2=3a2k+12③.由①③可得a2k+2a2k=12,由①②可得a2k+1a2k1=13,則數(shù)列a2k,a2k-1都是等差數(shù)列,首項分別為a2=74,a1=12,公差分別為12,13.則S2n=(a1+a3+a5+…+a2n1)+(a2+a4+…+a2n)=na1+n(n-1)2×13+na課時作業(yè)(三十)1.D[解析]由題意,得(23)a=22,解得a=4(2+3),故選D.2.A[解析]由題意得,q3=a6a3=124=18,則3.C[解析]設(shè)等比數(shù)列{an}的公比為q(q>0),由S3=14,a3=8,得S3=a1(1+q+q2)=14,a3=a1q2=8,可得a1=24.8[解析]因為a3a5a7=64,所以a53=64,解得a5=4,故a4=a5q5.1+5[解析]由等比數(shù)列的性質(zhì)知S2,S4S2,S6S4也成等比數(shù)列,所以(S4S2)2=S2·(S6S4),即(S42)2=2·(4S4),解得S4=1+5或S4=15(舍).6.B[解析]由a5=2S4+3,a6=2S5+3可得a6a5=2a5,則a6a5=3,7.B[解析]設(shè)等比數(shù)列{an}的公比為q.∵數(shù)列{an}為等比數(shù)列,且a3=4,a7=16,∴a52=a3·a7=(4)×(16)=64,又a5=a3q2=4q2<0,∴a5=8.故選8.A[解析]設(shè)等比數(shù)列{an}的公比為q.由log2a3+log2a10=1得log2a3a10=1,即a3a10=2.∵a5a6a8a9=16,∴(a5a8)(a6a7)q2=16,∴q2=4.由真數(shù)大于零得q>0,∴q=2.故選A.9.C[解析]由等比數(shù)列的性質(zhì)可知a5·a6=a4·a7=8,又a4+a7=2,故a4,a7是一元二次方程x22x8=0的兩個根,解得a4=2,a7=4或a4=4,a7=2,故a1=1,q3=2,a10=8或a1=8,q3=12,a10=1,所以a1+a10=710.A[解析]由各項均為正數(shù)的等比數(shù)列{an}滿足a7=a6+2a5,可得a1q6=a1q5+2a1q4,∴q2q2=0,∴q=2.∵aman=4a1,∴qm+n2=16,∴2m+n2=24,∴m+n=6,∴1m+4n=16(m+n)1m+4n=165+nm+4mn≥16(5+4)=32,當(dāng)且僅當(dāng)nm11.B[解析]∵等差數(shù)列{an}的公差d≠0,且a3,a5,a15成等比數(shù)列,a1=3,∴(3+4d)2=(3+2d)(3+14d),解得d=2或d=0,∵d≠0,∴d=2,則an=3+(n1)×(2)=52n,Sn=3n+n(n-1)2×(2)=4nn2,an·Sn=(52n)(4nn2)=2n313n2+20n.設(shè)f(x)=2x313x2+20x,則f'(x)=6x226x+20,令f'(x)=0,得x1=1,x2=103,則f(x)在1,103上單調(diào)遞減,在103,+∞上單調(diào)遞增.結(jié)合f(x)的單調(diào)性可知,當(dāng)n=3時,an·Sn取得最小值2×3313×32+20×3=3.故an·Sn的最小值為3.故選12.3[解析]由數(shù)列{Sn+2}也是等比數(shù)列可得S1+2,S2+2,S3+2成等比數(shù)列,則(S2+2)2=(S1+2)(S3+2),即(4+4q+2)2=(4+2)(4+4q+4q2+2),解得q=3或q=0(舍去).13.24(14100)[解析]由題得等比數(shù)列an的公比q=4a6a2=4116=12,所以a1=6,顯然數(shù)列anan+1也是等比數(shù)列,其首項為a1a2=18,公比q'=anan+1an-1an=q2=122=14,14.解:(1)由an+1=1+Sn得當(dāng)n≥2時,an=1+Sn1,兩式相減得an+1=2an.因為數(shù)列{an}是等比數(shù)列,所以a2=2a1,又因為a2=1+S1=1+a1,所以a1=1,則an=2n1.(2)易得數(shù)列l(wèi)g4002所以lg40020>lg40021>lg40022>…>lg4002由此可知當(dāng)n1=8,即n=9時,數(shù)列l(wèi)g400an的前n項和T15.解:(1)當(dāng)n=1時,a1=S1=λ.當(dāng)n≥2時,an=SnSn1=(n1)·2n(n2)·2n1=n·2n1.故數(shù)列an的通項公式為an=(2)由an·bn=n,可得bn=1因為數(shù)列bn為等比數(shù)列所以首項b1=1λ滿足n≥2的情況,故λ=1則Tn=b1+b2+…+bn=1-12n1-1因為Tn+1Tn=12n>0,所以Tn是遞增的,故Tn≥1且Tn<又存在m∈N*,使得m<Tn成立,則m的最大值為1.16.解:(1)設(shè)等比數(shù)列{an}的公比為q,由a3=32,S3=9得a1q2=32,a1(1+q+q2)=92,解得a1=6,q=12或a1=32則數(shù)列{an}的通項公式為an=32或an=6×-(2)當(dāng)an=32時,bn=log26a2n+1=log26×23=2由Tn=n2+105,得2n=n2+105,所以n=當(dāng)an=6×-12n-1時,bn=log26a2n故數(shù)列{bn}是首項為2,公差為2的等差數(shù)列,所以Tn=n2+n.由Tn=n2+105,得n2+n=n2+105,所以n=10或n=212(綜上知,n=70或10.課時作業(yè)(三十一)1.B[解析]由題知,所給數(shù)列的通項公式為an=2n+1+12n,則前n項和Sn=4(1-2n)1-2+122.A[解析]a1+a2+a3+a4+a5+a6+a7+a8+a9+a10=1+47+1013+1619+2225+28=5×3=15,故選A.3.C[解析]當(dāng)n為奇數(shù)時,n+1為偶數(shù),則an=n2(n+1)2=2n1,所以a1+a3+a5+a7=(3+7+11+15)=36.當(dāng)n為偶數(shù)時,n+1為奇數(shù),則an=n2+(n+1)2=2n+1,則a2+a4+a6+a8=5+9+13+17=44.所以a1+a2+…+a8=36+44=8,故選C.4.103[解析]an=13n+1+3n-2=13(3n+13n-2),則數(shù)列an的前40項和S40=13[(41)+(745.n2n+4[解析]設(shè)等差數(shù)列{an}的公差為d,由題意得a1+d=8,6a1+15d=66,解得a1=6,d=2,則an=2n+4,因此bn=2(n+1)(2n+4)6.C[解析]因為2n+12n=1+12n,所以Tn=n+112n,T10+1013=111210+1013=10241210,又n>T7.B[解析]設(shè)數(shù)列ann的公差為d,則a66=a1+5d=2+5d,a66=a33+3d=a612+3d,解得d=2,所以ann=a1+(n1)d=2n,an=2n2,所以S10=a1+a2a3+a4…a9+a10=2×12+2×222×32+2×42…2×92+2×102=2[(2212)+(4232)+…+(10292)]=2[(21)×(1+2)+(43)×(3+4)+8.C[解析]新數(shù)列為1×n2,12×n2,13×n2,…,1n×n2,所以a1a2+a2a3+a3a4+…+an1an=n2411×2+12×3+13×4+…+1(n-1)×n=n24112+1213+1314+…+9.A[解析]依題意有a2a1=3,a3+a2=5,a4a3=7,a5+a4=9,a6a5=11,a7+a6=13,a8a7=15,…,由此可得a1+a3=2,a5+a7=2,…,a2+a4=12,a6+a8=28,…,所以S32=(a1+a3+…+a31)+(a2+a4+…+a32)=8×2+8×12+8×72×16=560,故選A10.(n1)2n+1[解析]由題意得an=n,bn=2n1,則anbn=n·2n1,則數(shù)列anbn的前n項和Sn=1·20+2·21+3·22+…+n·2n1①,所以2Sn=1·21+2·22+3·23+…+(n1)·2n1+n·2n②.①②得Sn=1+21+22+…+2n1n·2n=1-2n1-2n·2n,整理得Sn=(11.1051[解析]設(shè)等差數(shù)列{an}的公差為d,因為a1+a5+2=0,所以2a1+4d+2=0,a1=12d.因為2S1,3S2,8S3成等比數(shù)列,所以16S1S3=9S22,即16(12d)(33d)=9(23d)2.因為d為整數(shù),所以解得d=2,則a1=3,所以an=32(n1)=52n.則1anan+1=1(5-2n)(3-2n)=1212n-512n-3,所以數(shù)列1anan+1的前10項和為112.1133[解析]當(dāng)n為奇數(shù)時,an+2=2an,故奇數(shù)項是以a1=1為首項,2為公比的等比數(shù)列;當(dāng)n為偶數(shù)時,an+2=an+2,故偶數(shù)項是以a2=2為首項,2為公差的等差數(shù)列,所以前20項中的奇數(shù)項和S奇=1-2101-2=2101=1023,前20項中的偶數(shù)項和S偶=10×2+10×92×2=110,所以S2013.解:(1)當(dāng)n=1時,a1=S1=2.由Sn=2n+12得Sn1=2n2(n≥2),∴an=SnSn1=2n+12n=2n(n≥2).當(dāng)n=1時,a1=2滿足上式,∴an=2n(n∈N*).(2)bn=1(2n+1)log222n-1+22n1=1(2n+1則Tn=12113+1315+…+12n-112n+1+(2+23+2=12112n+1+2(1-4n14.解:(1)因為a1=1,an+1an=2,所以an是首項為1,公差為2的等差數(shù)列所以an=1+(n1)×2=2n1.當(dāng)n=1時,b1=S1=2b1,所以b1=1.當(dāng)n≥2時,Sn=2bn①,Sn1=2bn1②,由①②得bn=bn+bn1,即bnbn所以bn是首項為1,公比為12的等比數(shù)列,故bn=(2)由(1)知cn=anbn=2n-12n-1,則Tn=120+312Tn=121+322+…+③④得12Tn=120+221+222+…+22n-12n-12n=1+1+1所以Tn=62n+315.A[解析]由a1=2,an+1=1+an1-an,得a2=3,a3=12,a4=13,a5=2,a6=3,…,由此可得數(shù)列{an}是以4為周期的周期數(shù)列,且a1a2a3a4=1,所以{an}的前2017項的積為a1a2a3a4…a2017=1×1×1×…×a1=2,所以{log2an}的前2017項的和為log2a1+log2a2+…+log2a2017=log2(a1a2…a16.(∞,2]∪[2,+∞)[解析]由題設(shè)可得an+1an=1nan+1n,即an+1=n+1nan+1n,即an+1n+1=ann+1n(n+1),所以an+1n+1ann=1n1n+1.令n=1,2,3,…,n可得a22a11=1112,a33a22=1213,a44a33=1314,…,an+1n+1ann=1n1n+1,累加得an+1n+1a11=11n課時作業(yè)(三十二)1.D[解析]根據(jù)韋達(dá)定理可得a2+a4=1,所以S5=5(a1+a5)22.A[解析]因為點A,B,C在一條直線上,所以a3+a2015=1,則S2017=2017(a1+a2017)23.C[解析]由an=ncos2nπ4sin2nπ4,得a1=0,a2=2,a3=0,a4=4,a5=0,a6=6,a7=0,…則a2n1=0,a2n=(1)n·2n,則S40=(a1+a3+…+a39)+(a2+a4+…+a40)=2+46+8…+40=2×10=20,故選4.8[解析]ann=12n+33n12,其中12n+33n≥212n·33n=66,當(dāng)且僅當(dāng)12n=33n即n=66時取等號.易知8<665.55987[解析]經(jīng)過10秒鐘后知道這條信息的人數(shù)為1+6,經(jīng)過20秒鐘后知道這條信息的人數(shù)為1+6+62,經(jīng)過30秒鐘后知道這條信息的人數(shù)為1+6+62+63,則經(jīng)過x個10秒鐘后知道這條信息的人數(shù)為1+6+62+…+6x,所以經(jīng)過一分鐘即經(jīng)過60秒鐘后知道此信息的人數(shù)為1+6+62+…+66=55987.6.D[解析]因為S1,S2,S4成等比數(shù)列,所以S22=S1·S4,即(2a12)2=a1(4a112),解得a1=1,故選D7.C[解析]由題意得公差d>0,且am>0,所以當(dāng)n>m時,Snan=SnSm+aman=am+am+1+…+an1>0,所以Sn>an,故選C.8.D[解析]設(shè)等比數(shù)列{an}的公比為q,則由a2,a4,a3成等差數(shù)列得,2a2q2=a2+a2q,即2q2q1=0,解得q=12或q=1(舍去).由a1a2a3a4a5=a35=11024=1210得a3=14=a1q2,所以a1=1,S5=9.C[解析]y'=2nxn1(n+1)xn,所以曲線y=(2x)xn在x=3處的切線的斜率為13n13n,所以切線方程為y=13n13n(x3)3n.令x=0,得an=(n+2)·3n,則ann+2=3n,所以數(shù)列ann+2的前n項和Sn=3(110.D[解析]由題意可得Sn+3=3×2n,Sn=3×2n3,由等比數(shù)列前n項和的特點可得數(shù)列an是首項為3,公比為2的等比數(shù)列,數(shù)列{an}的通項公式為an=3×2n1.設(shè)等比數(shù)列{bn}的公比為q,則b1qn1+b1qn=3×2n1,解得b1=1,q=2,數(shù)列bn的通項公式為bn=2n1,由等比數(shù)列的求和公式有Tn=2n1.則有Sn=3Tn,T