1⼀1*2*3+1⼀2*3*4+1⼀3*4*5......+1⼀18*19*20，计算过程

1/(n-1)(n+1)n=1/2[1/n(n-1)-1/n(n+1)]
所以原式=1/2[1/(2*1)-1/(2*3)]+1/2[1/(3*2)-1/(3*4)]+......+1/2[1/(19*18)-1/(19*20)]
=1/2[1/(2*1)-1/(2*3)+1/(3*2)-1/(3*4)+......+1/(19*18)-1/(19*20)]
=1/2[1/2-1/(19*20)]
=189/760

1/1*2*3+1/2*3*4+1/3*4*5......+1/18*19*20
=(1/1*2-1/2*3+1/2*3-1/3*4+1/3*4-1/4*5......+1/18*19-1/19*20)÷2
=(1/1*2-1/19*20)÷2
=(1/2-1/380)÷2
=189/380÷2
=189/760

1/(1*2*3)=(1/2)*(1/1*2-1/2*3)
1/(2*3*4)=(1/2)*(1/2*3-1/3*4)
1/(3*4*5)=(1/2)*(1/3*4-1/4*5)
以此类推，规律就出现了。

裂项相消
1/[n*(n+1)*(n+2)]
=(1/2)[1/n(n+1)-1/(n+1)(n+2)]
=(1/2)[1/n-2/(n+1)+1/(n+2)]
原式
=(1/2)[(1+1/2+……+1/18)-2*(1/2+1/3+……+1/19)+(1/3+1/4+……+1/20)]
=(1/2)[(1+1/2+1/19+1/20)-2*(1/2+1/19)]
=(1/2)[1-1/2-1/19+1/20]
=189/760