算1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100
问题描述：
算1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100
求,10点以前回答
答
1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100=（√2-1）/[(√2+1)(√2-1)]+(√3-√2)/[(√3+√2)(√3-√2)]+(√4-√3)/[(√4+√3)(√4-√3)]+……+(√100-√99)/[(√100+√99)(√100-√99)]=√2-1+...为何1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100=（√2-1）/[(√2+1)(√2-1)]+(√3-√2)/[(√3+√2)(√3-√2)]+(√4-√3)/[(√4+√3)(√4-√3)]+……+(√100-√99)/[(√100+√99)(√100-√99)]1/(√2+1)=(√2-1)/[(√2+1)(√2-1)]=(√2-1)/(2-1)=√2-11/(√3-√2)=(√3-√2)/[(√3+√2)(√3-√2)]=(√3-√2)/(3-2)=√3-√2……1/(√100+√99)=(√100-√99)/[(√100+√99)(√100-√99)]=(√100-√99)/(100-99)=√100-√99