分类：教育/科学-升学入学-高考
问题：S=1/1X2X3+1/2X3X4+...+1/n(n+1)(n? 

求和S=1/1x2x3+1/2x3x4+1/3x4x5+...+1x/n(n+1)(n+2)
(n=1,2,3,...)
回答：解.裂项法.
 1/[n(n+1)(n+2)]=(1/2){1/[n)n+1)]-1/[(n+1)(n+2)]}
 =(1/2)[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
 =(1/2)[1/n-2/(n+1)+1/(n+2)]
 S=1/1x2x3+1/2x3x4+1/3x4x5+...+1x/n(n+1)(n+2)
  =(1/2)[1/1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+/4-2/5+1/6
   +.....+1/n-2/(n+1)+1/(n+2)]
 =(1/2)[1-1/2-1/(n+1)+1/(n+2)]
 =(n^2+3n)/[4(n+1)(n+2)]