OCR FP1 2012 January — Question 4 6 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJanuary
Marks6
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Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyModerate -0.8 This is a straightforward application of standard summation formulae requiring expansion to Σr³ - 3Σr, then substitution of known formulae and factorisation. While it's Further Maths content, it's a routine algebraic exercise with no problem-solving or insight required, making it easier than average overall.
Spec4.06a Summation formulae: sum of r, r^2, r^3
4 Find \(\sum _ { r = 1 } ^ { n } r \left( r ^ { 2 } - 3 \right)\), expressing your answer in a fully factorised form.
Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{1}{4}n^2(n+1)^2 - \frac{3}{2}n(n+1)\)M1, DM1, A1 Express as difference of two series; Use standard series results; Obtain correct unsimplified answer
\(\frac{1}{4}n(n+1)(n+3)(n-2)\)M1, A1, A1 Attempt to factorise; At least factor of \(n(n+1)\); Obtain correct answer (from unsimplified answer)
[6] 
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{4}n^2(n+1)^2 - \frac{3}{2}n(n+1)$ | M1, DM1, A1 | Express as difference of two series; Use standard series results; Obtain correct unsimplified answer |
| $\frac{1}{4}n(n+1)(n+3)(n-2)$ | M1, A1, A1 | Attempt to factorise; At least factor of $n(n+1)$; Obtain correct answer (from unsimplified answer) |
| **[6]** | | |

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4 Find $\sum _ { r = 1 } ^ { n } r \left( r ^ { 2 } - 3 \right)$, expressing your answer in a fully factorised form.

\hfill \mbox{\textit{OCR FP1 2012 Q4 [6]}}
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Q1 4 Q2 5 Q3 6 Q4 6 Q5 6 Q6 6 Q7 9 Q8 8 Q9 10 Q10 12