1/(m+1)+2/(m+1)+…+m/(m+1)=(1+2+...+m)/(1+m)=m(1+m)/[2(1+m)]=m/2∴S(m)=1/2+1/3+2/3+...+1/(m+1)+...+m/(m+1)=1/2+2/2+3/2+...+m/2=(1+2+...+m)/2=m(1+m)/4项数n=1+2+...+k=[k(k+1)]/2当k=8时,n=36当k=9时,n=45...