Use the formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) to show that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 4 r - 3 ) = k n ( n + 1 ) \left( 2 n ^ { 2 } - 1 \right)$$
where \(k\) is a constant.
Hence evaluate
$$\sum _ { r = 20 } ^ { 40 } r ^ { 2 } ( 4 r - 3 )$$
(2 marks)