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某次复习课讲到这样一个问题:问题已知数列{an}的首项a1=2,前n项和为Sn,且满足2an=SnSn-1(n≥2,n∈N).求证:数列{1/Sn}成等差数列.证明因为当n≥2时,an=Sn-Sn-1.又2an=SnSn-1(n≥2,n∈N).所以2(Sn-Sn-1)=SnSn-1,故SSn-nSSn-n1-1=12,即S1n-S1n-1=-21(n≥2,n∈N).故{S1n}成等差数
A reviewing class addresses such a problem: Problem It is known that the first a1 = 2 of the sequence {an} and the first n terms are Sn and satisfy 2an = SnSn-1 (n≥2, n∈N). Proof: The sequence {1 / Sn} is an arithmetic sequence, which proves that for 2 n = Sn-Sn-1 and n = 2, n∈N when n≥2, 1) = SnSn-1, so SSn-nSSn-n1-1 = 12, that is, S1n-S1n-1 = -21 (n≥2, n∈N)