（新課標(biāo)）廣西2019高考數(shù)學(xué)二輪復(fù)習(xí) 專題對(duì)點(diǎn)練13 等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和.docx

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專題對(duì)點(diǎn)練13　等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和 1.已知各項(xiàng)都為正數(shù)的數(shù)列{an}滿足a1=1,an2-(2an+1-1)an-2an+1=0. (1)求a2,a3; (2)求{an}的通項(xiàng)公式. 2.(2018北京,文15)設(shè){an}是等差數(shù)列,且a1=ln 2,a2+a3=5ln 2. (1)求{an}的通項(xiàng)公式; (2)求ea1+ea2+…+ean. 3.(2018全國(guó)Ⅲ,文17)等比數(shù)列{an}中,a1=1,a5=4a3. (1)求{an}的通項(xiàng)公式; (2)記Sn為{an}的前n項(xiàng)和,若Sm=63,求m. 4.在等差數(shù)列{an}中,a2+a7=-23,a3+a8=-29. (1)求數(shù)列{an}的通項(xiàng)公式; (2)設(shè)數(shù)列{an+bn}是首項(xiàng)為1,公比為2的等比數(shù)列,求{bn}的前n項(xiàng)和Sn. 5.(2018天津,文18)設(shè){an}是等差數(shù)列,其前n項(xiàng)和為Sn(n∈N*);{bn}是等比數(shù)列,公比大于0,其前n項(xiàng)和為Tn(n∈N*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6. (1)求Sn和Tn; (2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整數(shù)n的值. 6.在等差數(shù)列{an}中,a7=8,a19=2a9. (1)求{an}的通項(xiàng)公式; (2)設(shè)bn=1nan,求數(shù)列{bn}的前n項(xiàng)和Sn. 7.已知{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,且a1+a2=6,a1a2=a3. (1)求數(shù)列{an}的通項(xiàng)公式; (2){bn}為各項(xiàng)非零的等差數(shù)列,其前n項(xiàng)和為Sn.已知S2n+1=bnbn+1,求數(shù)列bnan的前n項(xiàng)和Tn. 8.已知數(shù)列{an}是等差數(shù)列,其前n項(xiàng)和為Sn,數(shù)列{bn}是公比大于0的等比數(shù)列,且b1=-2a1=2,a3-b2=-1,S3-2b3=7. (1)求數(shù)列{an}和{bn}的通項(xiàng)公式; (2)設(shè)cn=(-1)n-1anbn,求數(shù)列{cn}的前n項(xiàng)和Tn.  專題對(duì)點(diǎn)練13答案 1.解 (1)由題意得a2=12,a3=14. (2)由an2-(2an+1-1)an-2an+1=0得2an+1(an+1)=an(an+1). 因?yàn)閧an}的各項(xiàng)都為正數(shù),所以an+1an=12. 故{an}是首項(xiàng)為1,公比為12的等比數(shù)列, 因此an=12n-1. 2.解 (1)設(shè)等差數(shù)列{an}的公差為d, ∵a2+a3=5ln 2, ∴2a1+3d=5ln 2. 又a1=ln 2,∴d=ln 2. ∴an=a1+(n-1)d=nln 2. (2)由(1)知an=nln 2. ∵ean=enln 2=eln 2n=2n, ∴{ean}是以2為首項(xiàng),2為公比的等比數(shù)列. ∴ea1+ea2+…+ean =2+22+…+2n =2n+1-2. ∴ea1+ea2+…+ean=2n+1-2. 3.解 (1)設(shè){an}的公比為q, 由題設(shè)得an=qn-1. 由已知得q4=4q2, 解得q=0(舍去),q=-2或q=2. 故an=(-2)n-1或an=2n-1. (2)若an=(-2)n-1,則Sn=1-(-2)n3.由Sm=63得(-2)m=-188,此方程沒有正整數(shù)解. 若an=2n-1,則Sn=2n-1.由Sm=63得2m=64,解得m=6. 綜上,m=6. 4.解 (1)設(shè)等差數(shù)列{an}的公差是d.由已知(a3+a8)-(a2+a7)=2d=-6,解得d=-3, ∴a2+a7=2a1+7d=-23,解得a1=-1, ∴數(shù)列{an}的通項(xiàng)公式為an=-3n+2. (2)由數(shù)列{an+bn}是首項(xiàng)為1,公比為2的等比數(shù)列, ∴an+bn=2n-1, ∴bn=2n-1-an=3n-2+2n-1, ∴Sn=[1+4+7+…+(3n-2)]+(1+2+22+…+2n-1)=n(3n-1)2+2n-1. 5.解 (1)設(shè)等比數(shù)列{bn}的公比為q.由b1=1,b3=b2+2,可得q2-q-2=0.因?yàn)閝>0,可得q=2,故bn=2n-1.所以,Tn=1-2n1-2=2n-1. 設(shè)等差數(shù)列{an}的公差為d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,從而a1=1,d=1,故an=n.所以,Sn=n(n+1)2. (2)由(1),有 T1+T2+…+Tn=(21+22+…+2n)-n=2(1-2n)1-2-n=2n+1-n-2. 由Sn+(T1+T2+…+Tn)=an+4bn可得,n(n+1)2+2n+1-n-2=n+2n+1, 整理得n2-3n-4=0,解得n=-1(舍),或n=4. 所以,n的值為4. 6.解 (1)設(shè)等差數(shù)列{an}的公差為d,則an=a1+(n-1)d. 因?yàn)閍7=8,所以a1+6d=8. 又a19=2a9,所以a1+18d=2(a1+8d), 解得a1=2,d=1,所以{an}的通項(xiàng)公式為an=n+1. (2)bn=1nan=1n(n+1)=1n-1n+1, 所以Sn=1-12+12-13+…+1n-1n+1=nn+1. 7.解 (1)設(shè){an}的公比為q,由題意知a1(1+q)=6,a12q=a1q2, 又an>0,解得a1=2,q=2,所以an=2n. (2)由題意知S2n+1=(2n+1)(b1+b2n+1)2=(2n+1)bn+1, 又S2n+1=bnbn+1,bn+1≠0,所以bn=2n+1. 令cn=bnan,則cn=2n+12n, 因此Tn=c1+c2+…+cn =32+522+723+…+2n-12n-1+2n+12n. 又12Tn=322+523+724+…+2n-12n+2n+12n+1,兩式相減得12Tn=32+12+122+…+12n-1-2n+12n+1, 所以Tn=5-2n+52n. 8.解 (1)設(shè)數(shù)列{an}的公差為d,數(shù)列{bn}的公比為q,q>0, ∵b1=-2a1=2,a3-b2=-1,S3-2b3=7, ∴a1=-1,-1+2d-2q=-1,3(-1)+3d-22q2=7,解得d=2,q=2. ∴an=-1+2(n-1)=2n-3,bn=2n. (2)cn=(-1)n-1anbn=(-1)n-1(2n-3)2n, ∴Tn=-12-122+323-524+…+(-1)n-2(2n-5)2n-1+(-1)n-1(2n-3)2n, 12Tn=-122-123+324+…+(-1)n-2(2n-5)2n+(-1)n-1(2n-3)2n+1, ∴32Tn=-12-12+122-123+…+(-1)n-112n-1+(-1)n-1(2n-3)2n+1=-12+-121--12n-11--12+(-1)n-1(2n-3)2n+1, ∴Tn=-59+29-12n-1+(-1)n-1(2n-3)32n.