已知数列{an}的前n项和为Sn,且对任意n∈N*,有2Sn=3an-2,则a1=______;Sn=______

2025-03-04 00:14:25
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回答1:

∵2Sn=3an-2,①
∴n=1时,2a1=3a1-2,解得a1=2.
n≥2时,2Sn-1=3an-1-2,②
①-②,得:2an=3an-3an-1
整理,得an=3an-1
an
an?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.