OCR FP1 2013 June — Question 5 6 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2013
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 (∑r³, ∑r², ∑r) that students memorize for FP1. It requires only direct substitution into known formulae and algebraic simplification/factorization, with no problem-solving or conceptual insight needed. While it's Further Maths content, it's a routine textbook exercise that's easier than average A-level questions overall.
Spec4.06a Summation formulae: sum of r, r^2, r^3
5 Find \(\sum _ { r = 1 } ^ { n } \left( 4 r ^ { 3 } - 3 r ^ { 2 } + r \right)\), giving your answer in a fully factorised form.
Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
\(4 \times \frac{1}{4}n^2(n+1)^2 - 3 \times \frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}n(n+1)\)M1 Express as sum of three series
A1Obtain 2 correct (unsimplified) terms
A1Obtain correct \(3^{rd}\) (unsimplified) term
\(n^3(n+1)\)M1 Attempt to factorise, at least factor of \(n\)
A2Obtain correct answer, A1 if not fully factorised
[6] 
## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $4 \times \frac{1}{4}n^2(n+1)^2 - 3 \times \frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}n(n+1)$ | M1 | Express as sum of three series |
| | A1 | Obtain 2 correct (unsimplified) terms |
| | A1 | Obtain correct $3^{rd}$ (unsimplified) term |
| $n^3(n+1)$ | M1 | Attempt to factorise, at least factor of $n$ |
| | A2 | Obtain correct answer, A1 if not fully factorised |
| **[6]** | | |

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

\hfill \mbox{\textit{OCR FP1 2013 Q5 [6]}}
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Q1 6 Q2 7 Q3 6 Q4 6 Q5 6 Q6 7 Q7 8 Q8 6 Q9 8 Q10 12