已知等差数列{an}的前n项和为Sn，令bn＝1/Sn，且a3b3＝1/2,S3＋S5＝21，求{bn}的通项公式及其前n项和Sn

用a代表a1
a3=a+2d
S3=3a+3d,所以b3=1/S3=(a+2d)/(3a+3d)=1/2
2a+4d=3a+3d
a=d

S3=3a+3d=6a
S5=5a+(5*4/2)d=5a+10d=5a+10a=15a
所以S3+S5=21a=21
a=d=1
所以Sn=na+[n(n-1)/2]d=n+n(n-1)/2=(n^2+n)/2
所以bn=1/Sn=2/(n^2+n)

bn=2/(n^2+n)=2/[n(n+1)]
所以S(bn)=2/(1*2)+2/(2*3)+……+2/[n(n+1)]
=2*{(1/1-1/2)+(1/2-1/3)+……+[1/n-1/(n+1)]}
=2*[1-1/(n+1)]
=2n/(n+1)