Standard +0.8 This is a Further Maths question requiring expansion of a cubic product, manipulation of standard summation formulae, and the telescoping technique to evaluate a sum from r=21 to 40. While the individual steps are methodical, the combination of algebraic manipulation, formula application, and the non-trivial range evaluation makes this moderately challenging for Further Maths students.
6. (a) Using the formulae for \(\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\), show that
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 3 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + k )$$
where \(k\) is a constant to be found.
(b) Hence evaluate \(\sum _ { r = 21 } ^ { 40 } r ( r + 1 ) ( r + 3 )\)
6. (a) Using the formulae for $\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$, show that
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 3 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + k )$$
where $k$ is a constant to be found.\\
(b) Hence evaluate $\sum _ { r = 21 } ^ { 40 } r ( r + 1 ) ( r + 3 )$\\
\hfill \mbox{\textit{Edexcel F1 Q6 [10]}}