`[(1)/(1*2)+(1)/(2*3)+(1)/(3*4)+....+(1)/(99*100)]`=?

[(1)/(1xx2)+(1)/(2xx3)+(1)/(3xx4)+,+(1)/(99xx100)]=

The value of (1)/(1xx2)+(1)/(2xx3)+(1)/(3xx4)+..... +(1)/(99xx100) is

Note the specialities of fractional numbers. It can be written as 1-1/2=(2-1)/(1xx2)=1/(1xx2) 1/2-1/3=(3-2)/(2xx3)=1/(2xx3) 1/3-1/4=(4-3)/(3xx4)=1/(3xx4) now answer the following. 1/(1xx2)+1/(2xx3)+1/(3xx4)+....(1)/(99xx100)= ____

[(1)/(1times2)+(1)/(2times3)+(1)/(3times4)+......+(1)/(99times100)] =

For a real number x,[x] denotes greatest integer function, then find value of [(1)/(2)]+[(1)/(2)+(1)/(100)]+[(1)/(2)+(2)/(100)]+...+[(1)/(2)+(99)/(100)]

For a real number x,[x] denotes the integral part of x. The value of [(1)/(4)]+[(1)/(4)+(1)/(00)]+[(1)/(4)+(2)/(100)]+[(1)/(4)+(3)/(100)]+.........+[(1)/(4)+(99)/(100)]=

[-(1)/(3)]+[-(1)/(3)-(1)/(100)]+[-(1)/(3)-(2)/(100)]+….+[-(1)/(3)-(99)/(100)] is equal to (where [.] denotes greatest integer function)