Standard +0.3 This is a straightforward algebraic manipulation question requiring expansion of the sum, application of two standard formulae (∑r and ∑r³), and simplification to match the given result. While it involves multiple steps and algebraic manipulation, it's a routine exercise in applying memorized formulae with no problem-solving insight required, making it slightly easier than average for Further Maths.
6 Using the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) show that
$$\sum _ { r = 1 } ^ { n } r \left( r ^ { 2 } - 3 \right) = \frac { 1 } { 4 } n ( n + 1 ) ( n + 3 ) ( n - 2 ) .$$
Correctly factorised to give fully factorised form
# Question 6:
$\sum_{r=1}^{n}\left[r(r^2-3)\right]=\sum_{r=1}^{n}r^3-3\sum_{r=1}^{n}r$ | M1 | Separate into separate sums (may be implied)
$=\frac{1}{4}n^2(n+1)^2-\frac{3}{2}n(n+1)$ | M1 | Substitution of standard result in terms of $n$
| A2 | For two correct terms (indivisible)
$=\frac{1}{4}n(n+1)(n(n+1)-6)$ | M1 | Attempt to factorise with $n(n+1)$
$=\frac{1}{4}n(n+1)(n^2+n-6)=\frac{1}{4}n(n+1)(n+3)(n-2)$ | A1 [6] | Correctly factorised to give fully factorised form
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6 Using the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ show that
$$\sum _ { r = 1 } ^ { n } r \left( r ^ { 2 } - 3 \right) = \frac { 1 } { 4 } n ( n + 1 ) ( n + 3 ) ( n - 2 ) .$$
\hfill \mbox{\textit{OCR MEI FP1 2009 Q6 [6]}}