设等差数列an的前n项和为Sn,已知bn=1/Sn,且a3*b3=1/2,S3+S5=21,求bn,bn前n项和Tn.

S3=3a1+3*2*d/2=3a1+3d
s5=5a1+5*4*d/2=5a1+10d
a3*b3=(a1+2d)/S3=(a1+2d)/(3a1+3d)
S3+S5=3a1+3d+5a1+10d=8a1+13d
(a1+2d)/(3a1+3d)=1/2
8a1+13d=21
解得,a1=1,d=1
Sn=na1+n(n-1)d/2=n(n+1)/2
bn=1/[n(n+1)/2]=2/[n(n+1)]
Tn=2/(1*2)+2/(2*3)+...+2/[n(n+1)]
=2*[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=2*[1-1/(n+1)]
=2n/(n+1)