(1/3)n(3n+2)= (1/3)n(3n+3) - (1/3)n= n(n+1) - n/3= (1/3)[ n(n+1)(n+2) - (n-1)n(n+1) ] - (1/6)[ n(n+1) - (n-1)n ] (1/3)n(3n+2)的前n项和=(1/3)n(n+1)(n+2) - (1/6)n(n+1)=(1/6)n(n+1)(2n+3)是1/(3n(3n+2））1/(3n(3n+2))这个没办法做用其他方法有没有以我所知没有，这是否你自己出的问题？