Standard +0.8 This is a Further Maths question requiring expansion of a cubic expression, manipulation of standard summation formulae, and then a non-trivial application involving the difference of two sums with shifted indices. Part (b) requires careful algebraic manipulation to reach the specified factored form. While systematic, it demands more sophistication than typical A-level questions and involves extended algebraic work across multiple steps.
5. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r ^ { 3 } , \sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } r ( r - 1 ) ( r - 3 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n - 1 ) ( 3 n - 10 )$$
(b) Hence show that
$$\sum _ { r = n + 1 } ^ { 2 n + 1 } r ( r - 1 ) ( r - 3 ) = \frac { 1 } { 12 } n ( n + 1 ) \left( a n ^ { 2 } + b n + c \right)$$
where \(a\), \(b\) and \(c\) are integers to be determined.
5. (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r ^ { 3 } , \sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } r ( r - 1 ) ( r - 3 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n - 1 ) ( 3 n - 10 )$$
(b) Hence show that
$$\sum _ { r = n + 1 } ^ { 2 n + 1 } r ( r - 1 ) ( r - 3 ) = \frac { 1 } { 12 } n ( n + 1 ) \left( a n ^ { 2 } + b n + c \right)$$
where $a$, $b$ and $c$ are integers to be determined.\\
\hfill \mbox{\textit{Edexcel F1 2021 Q5 [8]}}