B4=1/10,bn=1/Sn
S4=10 等差数列{an],设公差是d,那么 4A1+6d=10
S6-S3=15 6A1+15d-3A1-3d=3A1+12d=15
A1=1 d=1
所以An=n
Sn=n(n+1)/2
Bn=2/n(n+1)=2[1/n-1/(n+1)]
Tn=1/2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]=n/[2(n+1)]
B4=1/10,bn=1/Sn
S4=10 等差数列{an],设公差是d,那么 4A1+6d=10
S6-S3=15 6A1+15d-3A1-3d=3A1+12d=15
A1=1 d=1
所以An=n
Sn=n(n+1)/2
Bn=2/n(n+1)=2[1/n-1/(n+1)]
Tn=1/2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]=n/[2(n+1)]