- (a) Show that, for \(r > 0\)
$$\frac { 1 } { r ^ { 2 } } - \frac { 1 } { ( r + 1 ) ^ { 2 } } \equiv \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } }$$
(b) Hence prove that, for \(n \in \mathbb { N }\)
$$\sum _ { r = 1 } ^ { n } \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } } = \frac { n ( n + 2 ) } { ( n + 1 ) ^ { 2 } }$$
(c) Show that, for \(n \in \mathbb { N } , n > 1\)
$$\sum _ { r = n } ^ { 3 n } \frac { 6 r + 3 } { r ^ { 2 } ( r + 1 ) ^ { 2 } } = \frac { a n ^ { 2 } + b n + c } { n ^ { 2 } ( 3 n + 1 ) ^ { 2 } }$$
where \(a , b\) and \(c\) are constants to be found.