OCR MEI FP1 2005 January — Question 4 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2005
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyModerate -0.3 This is a straightforward application of standard summation formulae requiring expansion to Σr³ + 2Σr², then substitution of known formulae and factorisation. While it's a Further Maths question, it's a routine textbook exercise with no problem-solving insight needed, making it slightly easier than an average A-level question overall.
Spec4.06a Summation formulae: sum of r, r^2, r^3
4 Find \(\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r + 2 )\), giving your answer in a factorised form.
Question 4:
AnswerMarks Guidance
AnswerMark Guidance
\(\sum_{r=1}^{n} r^2(r+2) = \sum_{r=1}^{n} r^3 + 2\sum_{r=1}^{n} r^2\)M1, A1 Separate sums
\(= \frac{1}{4}n^2(n+1)^2 + \frac{1}{3}n(n+1)(2n+1)\)M1, A1 Use of formulae. Follow through from incorrect expansion in line 1
\(= \frac{1}{12}n(n+1)[3n(n+1) + 4(2n+1)]\)M1 Factorising
\(= \frac{1}{12}n(n+1)(3n^2+11n+4)\)A1
[6] 
# Question 4:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{n} r^2(r+2) = \sum_{r=1}^{n} r^3 + 2\sum_{r=1}^{n} r^2$ | M1, A1 | Separate sums |
| $= \frac{1}{4}n^2(n+1)^2 + \frac{1}{3}n(n+1)(2n+1)$ | M1, A1 | Use of formulae. Follow through from incorrect expansion in line 1 |
| $= \frac{1}{12}n(n+1)[3n(n+1) + 4(2n+1)]$ | M1 | Factorising |
| $= \frac{1}{12}n(n+1)(3n^2+11n+4)$ | A1 | |
| **[6]** | | |

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

\hfill \mbox{\textit{OCR MEI FP1 2005 Q4 [6]}}
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