如题，简单的方法
參考答案:1/n(n+1)(n+2)
=1/n(n+2)-1/(n+1)(n+2)=(1/2)[1/n-1/(n+2)]-[1/(n+1)-1/(n+2)]
所以原式
＝(1/2)[1-1/3+1/2-1/4+1/3-1/5......+1/(n+1)-1/(n+2)]
+[1/2-1/3+1/3-1/4+.........+1/(n+1)-1/(n+2)]
＝(1/2)[1+1/2-1/(n+1)-1/(n+2)]+[1/2-1/(n+2)]
＝1-(4n+5)/(2n^2+5n+2)
