数列{an}首项a1=3通项an与前n项的Sn之间满足2an=SnS(n-1) (n>=2)求证{1/Sn}为等差数列并求d_数学解答

数列{an}首项a1=3通项an与前n项的Sn之间满足2an=SnS(n-1) (n>=2)求证{1/Sn}为等差数列并求d

人气：467 ℃　时间：2020-06-03 01:12:35

解答

an = Sn - S
2an = Sn S
2(Sn - S) = Sn S
两端同时除 2 Sn S
1/S - 1/Sn = 1/2
1/Sn - 1/S = -1/2
因此 {1/Sn} 是公差为 -1/2 的等差数列