已知等差数列｛an｝的前n项和为Sn,且a3＝5,S3＝9.（1）求数列｛an｝的通向公式及Sn.（2）令bn＝1/S...

设公差为d,因5=a3=a1+2d,9=S3=a1+4d,解得：d=2,a1=1
所以：an=1+2(n-1)=2n-1,Sn=(1+2n-1)*n/2=n^2
bn=1/((n+1)^2-1)=1/n(n+2)=(1/2)(1/n-1/(n+2))
故：Tn=(1/2){1-1/3+1/2-1/4+1/3-1/5+...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2))
=(1/2)(3/2-1/(n+1)-1/(n+2))