计算:-3/(1*2)+5/(2*3)-7/(3*4)+9/(4*5)-…+2005/(1002*1003)=

来源:学生作业帮助网 编辑:六六作业网 时间:2024/05/04 03:25:32
计算:-3/(1*2)+5/(2*3)-7/(3*4)+9/(4*5)-…+2005/(1002*1003)=计算:-3/(1*2)+5/(2*3)-7/(3*4)+9/(4*5)-…+2005/(1

计算:-3/(1*2)+5/(2*3)-7/(3*4)+9/(4*5)-…+2005/(1002*1003)=
计算:-3/(1*2)+5/(2*3)-7/(3*4)+9/(4*5)-…+2005/(1002*1003)=

计算:-3/(1*2)+5/(2*3)-7/(3*4)+9/(4*5)-…+2005/(1002*1003)=
-3/(1*2)+5/(2*3)-7/(3*4)+9/(4*5)-…+2005/(1002*1003)
=-(1+2)/(1*2)+(2+3)/(2*3)-(3+4)/(3*4)+(4+5)/(4*5)-…-(1001+1002)/(1001*1002)+(1002+10003)/(1002*1003)
=-(1/2+1)+(1/3+1/2)-(1/4+1/3)+(1/5+1/4)+.-(1/1002+1/1001)+(1/1003+1/1002)
=-1/2-1+1/3+1/2-1/4-1/3+1/5+1/4+...-1/1002-1/1001+1/1003+1/1002
=-1+1/1002(中间两项,两两相消,只留下两项-1,1/1002)
=-1001/1002