在等差数列{an]中,前n项和Sn,若bn=1/Sn,b4=1/10,S6-S3=15,求1、{bn}的通项公式,2、T

设等差数列{an}的首项=a1,公差=d,则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
联立解得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)

由S6-S3=15得a5=5
a1+a4=a2+a3=5
设公差为d，则2a1+3d=5
a1+4d=5
所以a1=1,d=1
Sn通项公式为：n*(n+1)/2
bn=2/[n(n+1)]
Tn=2[1/1*2+1/2*3+...+1/n(n+1)]
=2[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=2-2/(n+1)
=2n/(n+1)

S6-S3=a4+a5+a6=15→a5=15/3=5
S5=S4+a5=10+5=15→a3=15/5=3
an=n
Sn=(n+1)n/2
bn=2/〔(n+1)n〕
Tn=2n/(n+1)