如果有有理数a,b满足a=2,b=1,求ab/1+（a+1）（b+1)/1+(a+2)(b+2)/1`````+(a+2007)(b+2007）/1的值
问题描述：
如果有有理数a,b满足a=2,b=1,求ab/1+（a+1）（b+1)/1+(a+2)(b+2)/1`````+(a+2007)(b+2007）/1的值
答
因1/[(a+n)(b+n)]=(a-b)[1/(b+n)]-[1/(a+n)]=1/(b+n)]-[1/(a+n)],(n=0,1,2,3,...),故原式=[1+(1/2)+(1/3)+...+(1/2007)+(1/2008)]-[(1/2)+(1/3)+(1/4)+...+(1/2008)+(1/2009)]=1-(1/2009)=2008/2009.