- Use the standard results for \(\sum _ { r = 1 } ^ { n } r\) and for \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) to show that, for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } r \left( r ^ { 2 } - 3 \right) = \frac { n } { 4 } ( n + a ) ( n + b ) ( n + c )$$
where \(a\), \(b\) and \(c\) are integers to be found.
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