Show that, for all integers \(r\),
$$\frac { 1 } { 2 r - 1 } - \frac { 1 } { 2 r + 1 } = \frac { 2 } { ( 2 r - 1 ) ( 2 r + 1 ) }$$
7
Hence, using the method of differences, show that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) } = \frac { a n } { b n + c }$$
where \(a\), \(b\) and \(c\) are integers to be determined.
7