> 数学 >
设数列{an}的前n项和为Sn,Sn=
a1(3n−1)
2
(对于所有n≥1),且a4=54,则a1的数值是______.
人气:260 ℃ 时间:2020-04-14 09:43:31
解答
设数列{an}的前n项和为Sn,Sn=
a1(3n−1)
2
(对于所有n≥1),
则a4=S4-S3=
a1(81−1)
2
a1(27−1)
2
=27a1

且a4=54,则a1=2
故答案为2
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