Given that \(\mathrm { f } ( r ) = \frac { 1 } { 4 r - 1 }\), show that
$$\mathrm { f } ( r ) - \mathrm { f } ( r + 1 ) = \frac { A } { ( 4 r - 1 ) ( 4 r + 3 ) }$$
where \(A\) is an integer.
Use the method of differences to find the value of \(\sum _ { r = 1 } ^ { 50 } \frac { 1 } { ( 4 r - 1 ) ( 4 r + 3 ) }\), giving your answer as a fraction in its simplest form. [0pt]
[4 marks]