OCR FP1 2011 June — Question 4 6 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks6
PaperDownload PDF ↗
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 splitting the sum, applying Σr² = n(n+1)(2n+1)/6, and algebraic manipulation to factorise. While it's Further Maths content, it's a routine textbook exercise with no problem-solving insight needed—just direct formula application and algebra.
Spec4.06a Summation formulae: sum of r, r^2, r^3
4 Find \(\sum _ { r = 1 } ^ { 2 n } \left( 3 r ^ { 2 } - \frac { 1 } { 2 } \right)\), expressing your answer in a fully factorised form.
Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(3 \times \frac{1}{6} \times 2n(2n+1)(4n+1) - \frac{1}{2} \times 2n\)M1 Express as sum of two series
A1 A1Each term correct a.e.f.
M1Attempt to factorise
\(2n^2(4n+3)\)A2 [6] Completely correct answer, (A1 if one factor not found) 
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $3 \times \frac{1}{6} \times 2n(2n+1)(4n+1) - \frac{1}{2} \times 2n$ | M1 | Express as sum of two series |
| | A1 A1 | Each term correct a.e.f. |
| | M1 | Attempt to factorise |
| $2n^2(4n+3)$ | A2 **[6]** | Completely correct answer, (A1 if one factor not found) |

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

\hfill \mbox{\textit{OCR FP1 2011 Q4 [6]}}
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