用裂项相消法.
因为 3an=3/[n(n+3)]=1/n -1/(n+3)
所以3Sn=3a1+3a2+3a3+3a4+...+3an
=(1-1/4) +(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+...+1/n -1/(n+3)
=1+1/2+1/3 -1/(n+1) -1/(n+2)-1/(n+3)
=11/6 +(3n²+12n+11)/[(n+1)(n+2)(n+3)]
然后,两边除以3,整理即可.