数列{an}前n项和Sn,已知Sn=98an-43×3n 43,求和3S1 32...
解:∵Sn=98an-43×3n 43,①∴Sn-1=98an-1-43×3n-1 43,②①-②,得an=98an-98an-1-43×3n 43×3n-1,整理,得18an 4×3n-2=98an-1 4×3n-1=9(18an-1 4×3n-3),又a1=S1=98a1-43×3 43,解得a1=643,18a1 4×3-1=4,∴{18an 4×3n-2}是首项为4,公比为9的等比数列,∴18an 4×3n-2=4×9n-1,∴an=329(9n-3n),∴Sn=329[9(1-9n)1-9-3(1-3n)1-3]=4(9n-1)-163(3n-1)=4×9n-16×3n-1 43...全部
解:∵Sn=98an-43×3n 43,①∴Sn-1=98an-1-43×3n-1 43,②①-②,得an=98an-98an-1-43×3n 43×3n-1,整理,得18an 4×3n-2=98an-1 4×3n-1=9(18an-1 4×3n-3),又a1=S1=98a1-43×3 43,解得a1=643,18a1 4×3-1=4,∴{18an 4×3n-2}是首项为4,公比为9的等比数列,∴18an 4×3n-2=4×9n-1,∴an=329(9n-3n),∴Sn=329[9(1-9n)1-9-3(1-3n)1-3]=4(9n-1)-163(3n-1)=4×9n-16×3n-1 43,∴3nSn=3n4×9n-16×3n-1 43=14×3n-163 43n 1=143(13n 3n 1-4),当n=1时,3S1=964<316,当n≥2时,3n 1-4>2×3n,∴3nSn=143(13n 3n 1-4)<34•13n 1-4<38×13n,∴3S1 32S2 … 3nSn<38(13 132 … 13n)=38×13(1-13n)1-13=38×12(1-13n)=316(1-13n)<316.综上所述,3S1 32S2 … 3nSn<316.。
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