新教材適用2023-2024學(xué)年高中數(shù)學(xué)第一章數(shù)列復(fù)習(xí)課北師大版選擇性必修第二冊

第1課時(shí)數(shù)列課后訓(xùn)練鞏固提升A組1.若在數(shù)列{an}中,a1=1,an+1=an2-1(n∈N+),則a1+a2+a3+a4+a5=(A.-1 B.1 C.0 D.2答案:A2.某種細(xì)胞開始有2個(gè),1小時(shí)后分裂成4個(gè)并死去1個(gè),2小時(shí)后分裂成6個(gè)并死去1個(gè),3小時(shí)后分裂成10個(gè)并死去1個(gè)……按此規(guī)律進(jìn)行下去,6小時(shí)后細(xì)胞存活的個(gè)數(shù)是().A.33 B.65 C.66 D.129解析:設(shè)細(xì)胞數(shù)由先到后排列構(gòu)成數(shù)列{an},則a1=2,an+1=2an-1,∴an+1∴數(shù)列{an-1}是首項(xiàng)為1,公比為2的等比數(shù)列,∴an-1=1×2n-1.∴an=2n-1+1,∴a7=26+1=65.答案:B3.已知等比數(shù)列{an}的通項(xiàng)公式為an=2·3n-1,現(xiàn)把每相鄰兩項(xiàng)之間都插入兩個(gè)數(shù)(與an不相等),構(gòu)成一個(gè)新的數(shù)列{bn},那么162是新數(shù)列{bn}的().A.第5項(xiàng) B.第12項(xiàng)C.第13項(xiàng) D.第6項(xiàng)解析:∵an=2×3n-1,∴162=2×3n-1,解得n=5.又每相鄰兩項(xiàng)之間插入兩個(gè)數(shù),∴a5=b13.答案:C4.定義:稱np1+p2+…+pn為n個(gè)正數(shù)p1,p2,…,pn的“均倒數(shù)”,若數(shù)列{an}的前n項(xiàng)的“均倒數(shù)”為12A.an=2n-1 B.an=4n-1C.an=4n-3 D.an=4n-5解析:∵數(shù)列{an}的前n項(xiàng)的“均倒數(shù)”為12∴數(shù)列{an}的前n項(xiàng)和Sn=(2n-1)n.∴an=Sn-Sn-1=(2n-1)n-(2n-3)(n-1)(n≥2).∴an=4n-3(n≥2),當(dāng)n=1時(shí),a1=S1=1,滿足an=4n-3.∴an=4n-3(n∈N+).答案:C5.(多選題)已知{an}是等比數(shù)列,下列說法中錯(cuò)誤的是().A.若a1<a2,則a4<a5B.若a1<a2,則a3<a4C.若S3>S2,則a1<a2D.若S3>S2,則a1>a2解析:等比數(shù)列{an}中,q2>0,當(dāng)a1<a2時(shí),可得a1q2<a2q2,即a3<a4,故B正確;但a4=a1q3和a5=a2q3不能判斷大小(因?yàn)閝3的正負(fù)不確定),故A錯(cuò)誤;當(dāng)S3>S2時(shí),a1+a2+a3>a1+a2,可得a3>0,即a1q2>0,可得a1>0,由于q不確定,不能確定a1,a2的大小,故C,D錯(cuò)誤.答案:ACD6.已知在等差數(shù)列{an}中,a1<a2<…<an,且a3,a6為x2-10x+16=0的兩個(gè)實(shí)根,則此數(shù)列的前n項(xiàng)和Sn=;通項(xiàng)公式為an=.

解析設(shè)等差數(shù)列{an}的公差為d,因?yàn)閍1<a2<…<an,所以公差d>0.因?yàn)閍3,a6為x2-10x+16=0的兩個(gè)實(shí)根,所以a3=2,a6=8,所以a解得a所以Sn=-2n+n(n-1)2×2=n2-3n,an=a1+(n-1)d=-2+(n-1)×答案:n2-3n2n-47.在正項(xiàng)等比數(shù)列{an}中,已知a3=4,a4=a2+6.(1)求{an}的前n項(xiàng)和Sn;(2)對于(1)中的Sn,設(shè)b1=S1,且bn+1-bn=Sn(n∈N+),求數(shù)列{bn}的通項(xiàng)公式.解:(1)設(shè)正項(xiàng)等比數(shù)列{an}的公比為q(q>0),由a3=4及a4=a2+6得,4q=4q+化簡得2q2-3q-2=0,解得q=2或q=-12(舍去負(fù)值)于是a1=4q2=1,所以Sn=1-2n1-2=2(2)由已知b1=S1=1,bn+1-bn=Sn=2n-1(n∈N+),所以當(dāng)n≥2時(shí),由累加法得bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1=(2n-1+2n-2+…+21)-(n-1)+1=2(1-2n-1又b1=1也適合上式,所以{bn}的通項(xiàng)公式為bn=2n-n,n∈N+.8.已知數(shù)列{an}是遞增的等比數(shù)列,且a1+a4=9,a2a3=8.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,bn=an+1SnSn+1,求數(shù)列{bn解:(1)由題設(shè)得a1a4=a2a3=8,結(jié)合a1+a4=9,可解得a1=1,a記公比為q.由a4=a1q3得q=2,故an=a1qn-1=2n-1.(2)Sn=a1(1-qn因?yàn)閎n=an所以Tn=b1+b2+…+bn=1S1-1S2+1S2-1S3+…+1SB組1.已知數(shù)列{an}的前n項(xiàng)和是Sn,如果Sn=3+2an(n∈N+),那么這個(gè)數(shù)列一定是().A.等比數(shù)列B.等差數(shù)列C.除去第一項(xiàng)后是等比數(shù)列D.除去第一項(xiàng)后是等差數(shù)列解析:由題意知,a1=3+2a1,即a1=-3.∵Sn=2an+3,an=Sn-Sn-1=(2an+3)-(2an-1+3)(n≥2),∴an=2an-1(n≥2),∴anan-1=∴數(shù)列{an}是等比數(shù)列,公比為2.答案:A2.設(shè)數(shù)列{an}是等比數(shù)列(公比q≠-1),它的前n項(xiàng)和、前2n項(xiàng)和及前3n項(xiàng)和分別為X,Y,Z,則下列等式恒成立的是().A.X+Z=2YB.Y(Y-X)=Z(Z-X)C.Y2=XZD.Y(Y-X)=X(Z-X)解析由題意,設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,則Sn=X,S2n=Y,S3n=Z.因?yàn)閿?shù)列{an}是等比數(shù)列(q≠-1),所以Sn,S2n-Sn,S3n-S2n成等比數(shù)列,即X,Y-X,Z-Y成等比數(shù)列,所以(Y-X)2=X·(Z-Y).整理,得Y2-XY=ZX-X2,即Y(Y-X)=X(Z-X).答案D3.(多選題)已知在數(shù)列{an}中,a1=1,an+1=2anan+2(n∈N+),A.{an}是等差數(shù)列 B.{an}是遞減數(shù)列C.1anD.1a解析:因?yàn)閍n+1=2anan+2,a1=1,所以1a即1a又a1=1,則1a1所以1an是以1為首項(xiàng),12為公差的等差數(shù)列所以1an=1a1+所以an=2n所以{an}不是等差數(shù)列,但為遞減數(shù)列.答案:BCD4.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a1>0,若存在自然數(shù)p≥10,使得Sp=ap,則當(dāng)n>p時(shí),Sn與an的大小關(guān)系是().A.an>Sn B.an≥SnC.an<Sn D.an≤Sn解析:設(shè)數(shù)列{an}的公差為d.∵Sp=ap,∴Sp-ap=0,即Sp-1=0.∵a1>0,∴d<0.Sn=d2n2+a1-d2n,an=dn+(a1-d).作出y=d2x2+a1-d2x,y=dx+(a(第4題)由圖象可知,當(dāng)n>p時(shí),an>Sn.答案:A5.已知公差不為零的正項(xiàng)等差數(shù)列{an}中,Sn為其前n項(xiàng)和,lga1,lga2,lga4也成等差數(shù)列,若a5=10,則S5=.

解析:設(shè){an}的公差為d,則d≠0.由lga1,lga2,lga4也成等差數(shù)列,得2lga2=lga1+lga4,∴a22=a1a即(a1+d)2=a1(a1+3d),d2=a1d.又d≠0,故d=a1,∴a5=5a1=10,∴d=a1=2,∴S5=5a1+5×42×答案:306.已知{an}是整數(shù)組成的數(shù)列,a1=1,且點(diǎn)(an,an+1)(n∈N+)在函數(shù)y=x2+1的圖象上(1)求數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列{bn}滿足b1=1,bn+1=bn+2an,求證:bnbn+2<(1)解:由已知得an+1=an+1,∴數(shù)列{an}是以1為首項(xiàng),公差為1的等差數(shù)列,∴an=1+(n-1)·1=n.(2)證明bn+1-bn=2an=2n,bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1=2n-1+2n-2+2n-3+…+2+1=1-2n1-2=當(dāng)n=1時(shí)也符合,∴bn=2n-1.∵bnbn+2-bn+12=(2n-1)(2n+2-1)-(2n+1-1)2=-5·2n+4·2n=-2n<0,∴bnbn+27.已知等差數(shù)列{an}滿足a3=7,a5+a7=26,等差數(shù)列{an}的前n項(xiàng)和為Sn.(1)求an及Sn;