Use standard results from the list of formulae (MF19) to show that
$$\sum _ { r = 1 } ^ { n } \left( 6 r ^ { 2 } + 6 r - 5 \right) = a n ^ { 3 } + b n ^ { 2 } + c n$$
where \(a\), \(b\) and \(c\) are integers to be determined.
Use the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 6 r ^ { 2 } + 6 r - 5 } { r ^ { 2 } + r }\) in terms of \(n\).
Find also \(\sum _ { r = n + 1 } ^ { 2 n } \frac { 6 r ^ { 2 } + 6 r - 5 } { r ^ { 2 } + r }\) in terms of \(n\).