Use the formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that \(\sum _ { r = 1 } ^ { n } ( 6 r - 3 ) ^ { 2 } = 3 n \left( 4 n ^ { 2 } - 1 \right)\).
Hence express \(\sum _ { r = 1 } ^ { 2 n } r ^ { 3 } - \sum _ { r = 1 } ^ { n } ( 6 r - 3 ) ^ { 2 }\) as a product of four linear factors in terms of \(n\). [0pt]
[4 marks]