1/x(x+3)+1/(x+3)(x+6)+...+1/(x+15)(x+18)

1/x(x+3)+1/(x+3)(x+6)+...+1/(x+15)(x+18)
=1/3[1/x-1/(x+3)]+1/3[1/(x+3)-1/(x+5)]+……+1/3[1/(x+15)-1/(x+18)]
=1/3[1/x-1/(x+3)+1/(x+3)-1/(x+5)+……+1/(x+15)-1/(x+18)]
=1/3[1/x-1/(x+18)]
=1/3[(x+18)/x(x+18)-x/x(x+18)]
=1/3[18/x(x+18)]
=6/x(x+18)

每一项分母中的两项都相差3
而：1/x(x+3)=(1/3)[(1/x)-(1/x+3)]
后面各项均可以分别为类似
所以，原式=(1/3)*{[(1/x)-(1/x+3)]+[(1/x+3)-(1/x+6)]+……+[(1/x+15)-(1/x+18)]}
=(1/3)*[(1/x)-(1/x+18)]
=(1/3)*[(x+18-x)/x*(x+18)]
=6/[x*(x+18)]