数列{an}中,a1=3且2an=SnS(n-1)(n≥2).(1)求证:{1/Sn}是等差数列.求公差.(2)求{an
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数列{an}中,a1=3且2an=SnS(n-1)(n≥2).(1)求证:{1/Sn}是等差数列.求公差.(2)求{an}通式
(1) an = Sn - S(n-1)
2(Sn-S(n-1)) = SnS(n-1)
Sn = 2S(n-1) / (2-S(n-1))
1/Sn = 1/S(n-1) - 1/2
所以 {1/Sn}为 首项是1/3,公差是 -1/2 的等差数列
(2)1/Sn 的通项公式是 1/Sn = 1/3 + (-1/2)*(n-1) = -1/2 n + 5/6
Sn = 6/(5-3n)
an = 1/2 ( 6/(5-3n) * 6/(8-3n)),n>=2
2(Sn-S(n-1)) = SnS(n-1)
Sn = 2S(n-1) / (2-S(n-1))
1/Sn = 1/S(n-1) - 1/2
所以 {1/Sn}为 首项是1/3,公差是 -1/2 的等差数列
(2)1/Sn 的通项公式是 1/Sn = 1/3 + (-1/2)*(n-1) = -1/2 n + 5/6
Sn = 6/(5-3n)
an = 1/2 ( 6/(5-3n) * 6/(8-3n)),n>=2
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