计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=_数学解答

计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/2x+4020

人气：209 ℃　时间：2020-01-27 23:54:23

解答

应该是1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/（2x+4020 ）吧1/(x+1)-1/(x+2)+1/(x+2）-1/(x+3)+1/(x+3)-1/(x+4)+...+1/(x+2009)-1/(x+2010)=1/（2x+4020） 1/(x+1)-1/(x+2010)=1/[2（x+...