列项相消法Sn=1/1·4 + 1/4·7 +...+1/(3n-2)(3n+1) =(1/3)*(1-1/4)+(1/3)*(1/4-1/7)+...+(1/3)* (1/(3n-2)-1/(3n+1))=(1/3)(1-1/4+1/4-1/7+...+1/(3n-2)-1/(3n+1)=(1/3)(1-1/(3n+1))=1/3 * 3n/(3n+1)=n/(3n+1)