Moderate -0.3 This is a straightforward application of standard summation formulae requiring expansion to r³ - r², then substitution of known formulae and algebraic simplification. While it's Further Maths content, it's a routine textbook exercise with no conceptual difficulty beyond remembering and applying the standard formulae, making it slightly easier than an average A-level question.
Attempt to use at least one standard result appropriately; Correct
\(= \tfrac{1}{12}n(n+1)(3n^2 - n - 2)\), or \(\tfrac{1}{12}n(n-1)(3n^2+5n+2)\)
M1 dep*
Attempt to factorise using either \(n(n-1)\) or \(n(n+1)\)
\(= \tfrac{1}{12}n(n+1)(n-1)(3n+2)\)
A2
All correct; SC A1 correct but \((3kn+2k)/12k\) seen
[6]
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n} r^2(r-1) = \sum_{r=1}^{n} r^3 - \sum_{r=1}^{n} r^2$ | M1* | Attempt to split into two summations |
| $= \tfrac{1}{4}n^2(n+1)^2 - \tfrac{1}{6}n(n+1)(2n+1)$ | M1, A1 | Attempt to use at least one standard result appropriately; Correct |
| $= \tfrac{1}{12}n(n+1)(3n^2 - n - 2)$, or $\tfrac{1}{12}n(n-1)(3n^2+5n+2)$ | M1 dep* | Attempt to factorise using either $n(n-1)$ or $n(n+1)$ |
| $= \tfrac{1}{12}n(n+1)(n-1)(3n+2)$ | A2 | All correct; SC A1 correct but $(3kn+2k)/12k$ seen |
| **[6]** | | |
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4 Using the standard summation formulae, find $\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r - 1 )$. Give your answer in a fully factorised form.
\hfill \mbox{\textit{OCR MEI FP1 2012 Q4 [6]}}