Use standard results from the list of formulae (MF19) to find \(\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } + 3 r + 1 \right)\) in terms of \(n\), simplifying your answer.
Show that
$$\frac { 1 } { r ^ { 3 } } - \frac { 1 } { ( r + 1 ) ^ { 3 } } = \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }$$
and hence use the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }\).
Deduce the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }\).