www.1862.net > 已知数列{An},Sn是其前n项的和,且满足3An=2Sn+n...

已知数列{An},Sn是其前n项的和,且满足3An=2Sn+n...

(1)∵3an=2Sn+n,∴a1=1,当n≥2时,3(an-an-1)=2an+1,即an=3an-1+1,∴an+12=3an-1+1+12=3(an-1+12),∴数列{an+12}是首项为32,公比为3的为等比数列;(2)由(1)知,an+12=32?3n-1,∴an=12×3n-12,∴Sn=a1+a2+…+an=12?3(1?3n)1?3-n2=34?3n-...

解:令bn=an+12 (1)3an=2sn+n,带入a1,有 3a1=2a1+1,得a1=1; 令n=2,带入:3a2=2(a1+a2)+2,得a2=4 令n=3,带入:3a3=2(a1+a2+a3)+3,得a3=13 b1*b3=(1+12)*(13+12)=325 b2*b2=(4+12)^2=256≠B1*B3,所以数列{an+12}并不是等比数列。 (...

解: n=1时,2a1=2S1=3a1-6 a1=6 n≥2时, 2an=2Sn-2S(n-1)=3an-6n-[3a(n-1)-6(n-1)] an=3a(n-1)+6 an+3=3a(n-1)+9=3[a(n-1)+3] (an+3)/[a(n-1)+3]=3,为定值 a1+3=6+3=9,数列{an+3}是以9为首项,3为公比的等比数列 an+3=9·3ⁿ⁻¹...

(I)∵2Sn=3an-1①∴2Sn-1=3an-1-1,(n≥2)②①-②得2Sn-2Sn-1=3an-3an-1=2an,即an=3an-1, 又n=1时,2S1=3a1-1=2a1∴a1=1∴{an}是以a1=1为首项,以q=3为公比的等比数列.∴an=a1qn-1=3n-1(II)Tn=1?30+2?31+3?32+…+n?3n-1,3Tn=1?31+2?32+3?33+…+n?3...

∵3an+1+2Sn=3,∴n≥2时,3an+2Sn-1=3,两式相减可得3an+1-an=0,∵a1=1,∴数列{an}是以1为首项,13为公比的等比数列,∴an=(13)n?1.故答案为:an=(13)n?1.

解(1)当n=1时,由题意可得6a1=a12+3a1+2∴a1=1或a1=2当n≥2时,6Sn=an2+3an+2,6Sn-1=an-12+3an-1+2,两式相减可得(an+an-1)(an-an-1-3)=0由题意可得,an+an-1>0∴an-an-1=3当a1=1时,an=3n-2,此时a42=a2?a9成立当a1=2时,an=3n-1,此时a42...

(1)n=1时,2S1+3=3a1?a1=3,n≥2时,2Sn+3=3an,①2Sn-1+3=3an-1,②由①-②得2an=3an-3an-1,∴an=3an-1,∴数列{an}是首项a1=3,公比为3的等比数列,∴an=3n(n∈N*),∴b2=a1=3,b4=a1+4=7,∴d=2,∴b1=1,∴bn=2n-1.(2)anbn=(2n-1)?3n,则Tn=1×3...

(Ⅰ)当n≥2时,2an=2Sn-2Sn-1=3an-2n-3an-1+2(n-1)即n≥2时,an=3an-1+2从而有n≥2时,an+1=3(an-1+1),又2a1=2S1=3a1-2得a1=2,故a1+1=3,∴数列{1+an}是等比数列,an+1=3n,即an=3n?1.(Ⅱ)bn=anan+1+1=3n?13n+1=13?13n+1,则Tn=n3?(1...

∵2Sn=3an-2,①∴n=1时,2a1=3a1-2,解得a1=2.n≥2时,2Sn-1=3an-1-2,②①-②,得:2an=3an-3an-1,整理,得an=3an-1,∴anan?1=3,∴{an}是首项为2,公比为3的等比数列,Sn=2(1?3n)1?3=3n-1.故答案为:2,3n-1.

(I)∵a1=1,且3an+1+2sn=3(n∈N?)∴当n=1时,3a2+2a1=3,∴a2=13…(2分)∴当n=2时,3a3+2(a1+a2)=3,∴a3=19…(3分)∵3an+1+2sn=3①∴当n≥2时,3an+2sn-1=3 ②由①-②,得3an+1-3an+2an=0…(5分)∴an+1an=13(n≥2),又∵a1=1,a2=13,…(7分)∴数...

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