If $$\left( X \right) = \frac{1}{x} - \frac{1}{{x + 1}},$$    then what is the value of f(1) + f(2) + f(3) + . . . . . + f(10)?

Solution (By Examveda Team)

$$\eqalign{ & f\left( 1 \right) = 1 - \frac{1}{2} \cr & f\left( 2 \right) = \frac{1}{2} - \frac{1}{3} \cr & f\left( 3 \right) = \frac{1}{3} - \frac{1}{4} \cr & f\left( 4 \right) = \frac{1}{4} - \frac{1}{5} \cr & 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4}\,.\,.\,.\,.\,.\,\frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{11}} \cr & = 1 - \frac{1}{{11}} \cr & = \frac{{10}}{{11}} \cr} $$