20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：高考大題專項(xiàng)(三)-數(shù)列-【含解析】-資料下載頁
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【正文】 ≥2),所以an2+2an+1=an+12,即(an+1)2=an+12(n≥2).因?yàn)閿?shù)列{an}的各項(xiàng)均為正數(shù),所以當(dāng)n≥2時(shí),an+1=an+1.(2)解由(1)得a4=a2+2,a8=a2+6,因?yàn)閍4是a2與a8的等比中項(xiàng),所以a42=a2a8,即(a2+2)2=a2(a2+6),解得a2=2,又2a1+2=a22,所以a1=1,所以a2a1=1,從而an+1an=1對n∈N*恒成立,所以數(shù)列{an}是以1為首項(xiàng),1為公差的等差數(shù)列,所以an=n.所以2nan=n2n.所以Tn=12+222+…+(n1)2n1+n2n,2Tn=122+223+…+(n1)2n+n2n+1兩式相減,得Tn=2+22+…+2nn2n+1=2(12n)12n2n+1=(1n)2n+12,所以Tn=(n1)2n+1+2.例6解(1)①當(dāng)λ=1時(shí),anSn+1an+1Sn=an+1an,則anSn+1+an=an+1Sn+an+1,即(Sn+1+1)an=(Sn+1)an+1.∵數(shù)列{an}的各項(xiàng)均為正數(shù),∴an+1an=Sn+1+1Sn+1.∴a2a1a3a2…an+1an=S2+1S1+1S3+1S2+1…Sn+1+1Sn+1,化簡,得Sn+1+1=2an+1,①∴當(dāng)n≥2時(shí),Sn+1=2an,②②①,得an+1=2an,∵當(dāng)n=1時(shí),a2=2,∴當(dāng)n=1時(shí),上式也成立,∴數(shù)列{an}是首項(xiàng)為1,公比為2的等比數(shù)列,即an=2n1.(2),令n=1,得a2=λ+1。令n=2,得a3=(λ+1)2.要使數(shù)列{an}是等差數(shù)列,必須有2a2=a1+a3,解得λ=0.當(dāng)λ=0時(shí),Sn+1an=(Sn+1)an+1,且a2=a1=1.當(dāng)n≥2時(shí),Sn+1(SnSn1)=(Sn+1)(Sn+1Sn),整理,得Sn2+Sn=Sn+1Sn1+Sn+1,即Sn+1Sn1+1=Sn+1Sn,從而S2+1S1+1S3+1S2+1…Sn+1Sn1+1=S3S2S4S3…Sn+1Sn,化簡,得Sn+1=Sn+1,即an+1=1.綜上所述,可得an=1,n∈N*.∴當(dāng)λ=0時(shí),數(shù)列{an}是等差數(shù)列.對點(diǎn)訓(xùn)練6解(1)∵a1a2a3=2b3,a1a2=2b2,b3b2=3,∴a3=2b32b2=2b3b2=8,又a1=2,∴8=2q2,∴q2=4,解得q=2或q=2,∵a1a2a3…an=2bn0(n∈N*),故舍去q=2,∴an=2n,∴a1a2a3…an=2(1+2+3+…+n)=2n(n+1)2,∴bn=n(n+1)2.(2)(1)得,=n+1n=1+1n,假設(shè)存在正整數(shù)m,n(m≠n),使c2,cm,成等差數(shù)列,則2cm=c2+,即21+1m=32+1+1n,∴2m=12+1n,故n=2m4m,∵n0,m0,∴0m4,∴m=2,n=2(舍去)或m=3,n=6,所以存在正整數(shù)m=3,n=6,使c2,cm,成等差數(shù)列.9
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