設(shè)等差數(shù)列{an}的首項(xiàng)=a1,公差=d,則Sn=na1+n(n-1）*d/2
∴S6-S3=(6a1+15d）-（3a1+3d)=3a1+12d=15,
b4=1/10=1/S4=1/(4a1+6d）,∴4a1+6d=10
聯(lián)立解得a1=d=1
∴Sn=n+n(n-1)/2=n(n+1)/2
bn=1/Sn=2/n(n+1),
∵bn=2/n(n+1）=2[1/n-1/(n+1)],
∴Tn=b1+b2+b3+……+bn
=2[1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)