已知等差数列{an}的前n项和为Sn,bn=1/Sn,且a3b3=1/2,S3+S5=21.1、求{bn}的通项公式 2、若BA*BC＝3/2,求a+c的值
问题描述：
已知等差数列{an}的前n项和为Sn,bn=1/Sn,且a3b3=1/2,S3+S5=21.
1、求{bn}的通项公式
2、若BA*BC＝3/2,求a+c的值
答
b3=1/S3=1/(3a1+3d)
a3=a1+2d
S3+S5=3a1+3d+5a1+10d=8a1+13d------①
(a1+2d)/(3a1+3d)=1/2--------------②
联立①②,得
a1=d=1
an=n,Sn=n(n+1)/2
bn=2/n(n+1)
答
b3=1/S3所以a3b3=a3/S3=1/2得S3=2a3,S5=5a3所以S3+S5=7a3=21,a3=3又S3=3a2=2a3=6,a2=2所以an的公比是1所以an=n,Sn=n(n+1)/2所以bn=2/n(n+1)
答
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...