已知等差数列{an}的前n项和为Sn,bn=1/Sn,且a3·b3=1/2,S3+S5=21
问题描述：
已知等差数列{an}的前n项和为Sn,bn=1/Sn,且a3·b3=1/2,S3+S5=21
①求数列{bn}的通项公式
②求证b1+b2+...+bn
答
a(n) = a1 + (n-1)d,S(n) = na1 + n(n-1)d/2.b(n) = 1/S(n) = 1/[na1 + n(n-1)d/2] = 2/[2na1 + n(n-1)d],1/2 = a3b3 = [a1+2d]*2/[6a1+6d] = (a1+2d)/(3a1+3d),3a1+3d = 2a1 + 4d,a1=d.b(n) = 2/[2na1 + n(n-1)d] =...