1/1+根號(hào)2+1/根號(hào)2+根號(hào)3+1/根號(hào)3+根號(hào)4……+1/根號(hào)99+根號(hào)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+...為何1/1+根號(hào)2+1/根號(hào)2+根號(hào)3+1/根號(hào)3+根號(hào)4……+1/根號(hào)99+根號(hào)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