- (a) Use the standard results for summations to show that, for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } r \left( 2 r ^ { 2 } - 3 r - 1 \right) = \frac { 1 } { 2 } n ( n + 1 ) ^ { 2 } ( n - 2 )$$
(b) Hence show that, for all positive integers \(n\),
$$\sum _ { r = n } ^ { 2 n } r \left( 2 r ^ { 2 } - 3 r - 1 \right) = \frac { 1 } { 2 } n ( n - 1 ) ( a n + b ) ( c n + d )$$
where \(a\), \(b\), \(c\) and \(d\) are integers to be determined.