题目
等差数列{An}的前n项和为Sn已知Bn=1/Sn且A3B3=1/2,S3+S5=21,求数列{Bn}的通项公式.
提问时间：2020-11-02
答案
a(n) = a + (n-1)d,n = 1,2,...
S(n) = na + n(n-1)d/2.
b(n) = 1/S(n) = 1/[na + n(n-1)d/2] = 2/[2na + n(n-1)d],
1/2 = a(3)b(3) = [a+2d]*2/[6a+6d] = (a+2d)/(3a+3d),
3a+3d = 2a + 4d,a = d.
b(n) = 2/[2na + n(n-1)d] = 2/[n(n+1)d].
S(n) = na + n(n-1)d/2 = nd[2 + n-1]/2 = n(n+1)d/2.
21 = S(3)+S(5) = 3*4d/2 + 5*6d/2 = 6d + 15d = 21d,d = 1,
b(n) = 2/[n(n+1)d] = 2/[n(n+1)],n = 1,2,...
S(n) = na + n(n-1)d/2.
b(n) = 1/S(n) = 1/[na + n(n-1)d/2] = 2/[2na + n(n-1)d],
1/2 = a(3)b(3) = [a+2d]*2/[6a+6d] = (a+2d)/(3a+3d),
3a+3d = 2a + 4d,a = d.
b(n) = 2/[2na + n(n-1)d] = 2/[n(n+1)d].
S(n) = na + n(n-1)d/2 = nd[2 + n-1]/2 = n(n+1)d/2.
21 = S(3)+S(5) = 3*4d/2 + 5*6d/2 = 6d + 15d = 21d,d = 1,
b(n) = 2/[n(n+1)d] = 2/[n(n+1)],n = 1,2,...
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