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