- (a) Prove by induction that for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$
(b) Use the standard results for \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } r ( r + 6 ) ( r - 6 ) = \frac { 1 } { 4 } n ( n + 1 ) ( n - 8 ) ( n + 9 )$$
(c) Hence find the value of \(n\) that satisfies
$$\sum _ { r = 1 } ^ { n } r ( r + 6 ) ( r - 6 ) = 17 \sum _ { r = 1 } ^ { n } r ^ { 2 }$$