- (a) Express
$$\frac { 1 } { ( 2 n - 1 ) ( 2 n + 1 ) ( 2 n + 3 ) }$$
in partial fractions.
(b) Hence, using the method of differences, show that for all integer values of \(n\),
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) } = \frac { n ( n + 2 ) } { a ( 2 n + b ) ( 2 n + c ) }$$
where \(a\), \(b\) and \(c\) are integers to be determined.