OCR MEI FP1 2008 January — Question 4 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyModerate -0.8 This is a straightforward algebraic manipulation question requiring expansion of (r+1)(r-2) into r² - r - 2, then applying standard summation formulae. It's routine bookwork for FP1 with clear signposting and no problem-solving insight needed, making it easier than average but not trivial since it requires careful algebraic manipulation across multiple steps.
Spec4.06a Summation formulae: sum of r, r^2, r^3
4 Using the standard formulae for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\), show that \(\sum _ { r = 1 } ^ { n } [ ( r + 1 ) ( r - 2 ) ] = \frac { 1 } { 3 } n \left( n ^ { 2 } - 7 \right)\).
Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\sum_{r=1}^{n}[(r+1)(r-2)] = \sum_{r=1}^{n}r^2 - \sum_{r=1}^{n}r - 2n\)M1 Attempt to split sum up
\(= \frac{1}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1) - 2n\)A2 Minus one each error
\(= \frac{1}{6}n\left[(n+1)(2n+1) - 3(n+1) - 12\right]\)M1 Attempt to factorise
\(= \frac{1}{6}n\left(2n^2 + 3n + 1 - 3n - 3 - 12\right)\)M1 Collecting terms
\(= \frac{1}{6}n\left(2n^2 - 14\right)\)
\(= \frac{1}{3}n\left(n^2 - 7\right)\)A1 [6] All correct 
# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}[(r+1)(r-2)] = \sum_{r=1}^{n}r^2 - \sum_{r=1}^{n}r - 2n$ | M1 | Attempt to split sum up |
| $= \frac{1}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1) - 2n$ | A2 | Minus one each error |
| $= \frac{1}{6}n\left[(n+1)(2n+1) - 3(n+1) - 12\right]$ | M1 | Attempt to factorise |
| $= \frac{1}{6}n\left(2n^2 + 3n + 1 - 3n - 3 - 12\right)$ | M1 | Collecting terms |
| $= \frac{1}{6}n\left(2n^2 - 14\right)$ | | |
| $= \frac{1}{3}n\left(n^2 - 7\right)$ | A1 **[6]** | All correct |

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4 Using the standard formulae for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$, show that $\sum _ { r = 1 } ^ { n } [ ( r + 1 ) ( r - 2 ) ] = \frac { 1 } { 3 } n \left( n ^ { 2 } - 7 \right)$.

\hfill \mbox{\textit{OCR MEI FP1 2008 Q4 [6]}}
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