1/(2根号1+1根号2)+1/(3根号2+2根号3)+.+1/(100根号99+99根号100)
问题描述：
1/(2根号1+1根号2)+1/(3根号2+2根号3)+.+1/(100根号99+99根号100)
答
1/(2+根号2)=(2-根号2)/(2+根号2)*(2-根号2)
=(2-根号2)/2
=1-(根号2)/2
1/(3根号2+2根号3)=(3根号2-2根号3)/(3根号2-2根号3)*(3根号2+2根号3) =(3根号2-2根号3)/6
=(根号2)/2 -(根号3)/3
同理,1/(100根号99+99根号100)=(根号99)/99 -(根号100)/100
所以, 1/(2根号1+1根号2)+1/(3根号2+2根号3)+.......+1/(100根号99+99根号100)
=[1-(根号2)/2]+[(根号2)/2-(根号3)/3]+......+[(根号99)/99 -(根号100)/100]
=1-(根号100)/100
=1-1/10=0.9
答
1/(2根号1+1根号2)=(2根号1-1根号2)/(4-2)=根号1-(根号2)/21/(3根号2+2根号3)=(3根号2-2根号3)/(18-12)=(根号2)/2-(根号3)/31/(2根号1+1根号2)+1/(3根号2+2根号3)+.+1/(100根号99+99根号100) =根号1-(根号)/2+(根号)...