1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2004)(x+2005)=1/x-1/(x+1)+1/(x+1)-1/(x+2)+-----+(1/(x+2004)-1/(x+2005)=1/x-1/(x+2005)=2005/(x^2+2005x)用裂项相消法1/（x+n)(x+n+1)=1/(x+n)-1/(x+n+1)