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)]
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