- (a) Express \(\frac { 1 } { r ( r + 2 ) }\) in partial fractions.
(b) Hence prove, by the method of differences, that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 2 ) } = \frac { n ( a n + b ) } { 4 ( n + 1 ) ( n + 2 ) }$$
where \(a\) and \(b\) are constants to be found.
(c) Hence show that
$$\sum _ { r = n + 1 } ^ { 2 n } \frac { 1 } { r ( r + 2 ) } = \frac { n ( 4 n + 5 ) } { 4 ( n + 1 ) ( n + 2 ) ( 2 n + 1 ) }$$