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设数列{an}的前n项和Sn=4/3an-2/3,求首项a1与通项an
人气:288 ℃ 时间:2020-05-26 21:41:45
解答
sn-1=4/3an-1-2/3
Sn-Sn-1=4/3an-4/3an-1
an=4/3an-4/3an-1
an/an-1=4
∴﹛an﹜为等比数列,公比为4,据题目原式得a1=2
∴an=2^(2n-1)
Sn LZ自己求吧,这种题型都是用Sn-Sn-1就可解出,很容易就出来了,希望多考虑.
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