Let $$f\left( n \right) = \left[ {\frac{1}{2} + \frac{n}{{100}}} \right]$$    where $$\left[ x \right]$$ denotes the integral part of $$x.$$ Then the value of $$\sum\limits_{n = 1}^{100} {f\left( n \right)} $$  is

Solution :
$$\eqalign{ & f\left( 1 \right) = \left[ {\frac{1}{2} + \frac{1}{{100}}} \right] = 0,....., \cr & f\left( {49} \right) = \left[ {\frac{1}{2} + \frac{{49}}{{100}}} \right] = \left[ {\frac{{99}}{{100}}} \right] = 0, \cr & f\left( {50} \right) = \left[ {\frac{1}{2} + \frac{{50}}{{100}}} \right] = 1, \cr & f\left( {51} \right) = \left[ {\frac{1}{2} + \frac{{51}}{{100}}} \right] = 1,....., \cr & f\left( {100} \right) = \left[ {\frac{1}{2} + \frac{{100}}{{100}}} \right] = 1. \cr} $$