1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]
=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+3)+1/(x+3)-1/(x+6)+.+1/(x+15)-1/(x+18)]
=1/x-1/(x+18)
=1/3*(x+18-x)/x(x+18)
=6/(x^2+18x)