- (a) Use the standard results for summations to show that, for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r + 1 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( a n + b )$$
where \(a\) and \(b\) are integers to be determined.
(b) Hence show that, for all positive integers \(k\),
$$\sum _ { r = k + 1 } ^ { 3 k } r ^ { 2 } ( r + 1 ) = \frac { 1 } { 3 } k ( 3 k + 1 ) \left( A k ^ { 2 } + B k + C \right)$$
where \(A , B\) and \(C\) are integers to be determined.
(c) Hence, using algebra and making your method clear, determine the value of \(k\) for which
$$25 \sum _ { r = k + 1 } ^ { 3 k } r ^ { 2 } ( r + 1 ) = 192 k ^ { 3 } ( 3 k + 1 )$$