
已知数列{an}的前n项和Sn,且满足a1=4/3,(4n-1)an=3×4n-1Sn,求数列{Sn}的通项公式
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(4n-1)an=3×(4n-1)Sn
4×(4n-1)Sn -(4n-1)(Sn-an)=0
4×(4n-1)Sn -(4n-1)S(n-1)=0
Sn/S(n-1)=1/4
Sn/S1= (1/4)^(n-1)
Sn= a1(1/4)^(n-1)
= (1/3)(1/4)^(n-2)
4×(4n-1)Sn -(4n-1)(Sn-an)=0
4×(4n-1)Sn -(4n-1)S(n-1)=0
Sn/S(n-1)=1/4
Sn/S1= (1/4)^(n-1)
Sn= a1(1/4)^(n-1)
= (1/3)(1/4)^(n-2)
更多追问追答
追问
那个3怎么没了啊?
追答
不好意思,
(4n-1)an=3×(4n-1)Sn
2×(4n-1)Sn +(4n-1)Sn= (4n-1)an
2×(4n-1)Sn +(4n-1)(Sn-an)=0
2×(4n-1)Sn +(4n-1)S(n-1)=0
Sn/S(n-1) = -1/2
Sn/S1= (-1/2)^(n-1)
Sn= a1(-1/2)^(n-1)
= (4/3)(-1/2)^(n-1)
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