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...

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。 (...

(Ⅰ)因为2Sn=3an-1,所以2Sn-1=3an-1-1,(n≥2)两式相减得2an=3an-3an-1,所以 an=3an-1,所以数列{an}是等比数列的公比q=3当n=1,得2a1=3a1-1,解得a1=1.则an=3n-1.(Ⅱ) bn=an+(-1)nlog3an=3n-1+(-1)nlog33n-1=3n-1+(-1)n(n-1),...

(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)由题意可得数列{an}是首项为1,公比为3的等比数列,故an=3n-1,Sn=1?3n1?3=3n?12;(Ⅱ)可得b1=a2=3,b3=a1+a2+a3=1+3+9=13,故数列{bn}的公差为12(13-3)=5故Tn=3n+n(n?1)2×5=5n2+n2

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...

LZ您好 解法一 2S[n-1]=3a[n-1]-3 将这个式子与已知条件两边相减(注意!!S[n]-S[n-1]=a[n]!!) 2a[n]=3a[n]-3-3a[n-1]+3 a[n]=3a[n-1] 所以当n≥2时,该数列是一个等比数列 当n=1时,2a[1]=3a[1]-3 a[1]=3 2(a[1]+a[2])=3a[2]-3 a[2]=9 可见a[2]/a[...

(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.

对3an+1+an=4 变形得:3(an+1-1)=-(an-1)即:an+1?1an?1=?13故可以分析得到数列bn=an-1为首项为8公比为?13的等比数列.所以bn=an-1=8×(?13)n?1an=8×(?13)n?1+1=bn+1所以Sn=Sbn+n=8[1?(?13)n]1?(?13)+n=6?6×(?13)n+n|Sn-n-6|=|?6×(?13)n|<...

解: n≥2时, 6an=6Sn-6S(n-1)=an²+3an+2-[a(n-1)²+3a(n-1)+2] an²-a(n-1)²-3an-3a(n-1)=0 [an+a(n-1)][an-a(n-1)-3]=0 数列是正项数列,an+a(n-1)恒>0,因此只有an-a(n-1)-3=0 an-a(n-1)=3,为定值 数列{an}是以3为公差的...

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