设 1/(n*(n 1)*(n 2))=A/n+B/(n+1)+C/(n+2)
所以,A(n+1)(n+2)+Bn(n+2)+Cn(n+1)=1
A(n^2+3n+2)+B(n^2+2n)+C(n^2+n)=1
(A+B+C)n^2+( )n+2A=1
得 A+B+C=0
3A+2B+C=0
2A=1
即 A=1/2
B=-1
C=1/2
1/(n*(n 1)*(n 2))=1/2n-1/(n+1)+1/2(n+2)