- (a) Express \(\frac { 4 r + 2 } { r ( r + 1 ) ( r + 2 ) }\) in partial fractions.
(b) Hence, using the method of differences, prove that
$$\sum _ { r = 1 } ^ { n } \frac { 4 r + 2 } { r ( r + 1 ) ( r + 2 ) } = \frac { n ( a n + b ) } { 2 ( n + 1 ) ( n + 2 ) }$$
where \(a\) and \(b\) are constants to be found.