OCR MEI FP1 2006 January — Question 3 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2006
SessionJanuary
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 expansion to r² - 1, then using Σr² and Σ1 formulae. While it's Further Maths content, it's a routine algebraic manipulation with no problem-solving insight needed, making it easier than average overall.
Spec4.06a Summation formulae: sum of r, r^2, r^3
3 Find \(\sum _ { r = 1 } ^ { n } ( r + 1 ) ( r - 1 )\), expressing your answer in a fully factorised form.
Question 3:
AnswerMarks Guidance
\(\sum_{r=1}^{n}(r+1)(r-1) = \sum_{r=1}^{n}(r^2 - 1)\)M1 expanding
\(= \frac{n(n+1)(2n+1)}{6} - n\)M1 A1 using standard result
\(= \frac{n(n+1)(2n+1) - 6n}{6}\)M1 combining over common denominator
\(= \frac{n[(n+1)(2n+1) - 6]}{6}\)A1 factoring \(n\)
\(= \frac{n(2n^2 + 3n - 5)}{6} = \frac{n(2n+5)(n-1)}{6}\)A1 fully factorised 
# Question 3:

$\sum_{r=1}^{n}(r+1)(r-1) = \sum_{r=1}^{n}(r^2 - 1)$ | M1 | expanding

$= \frac{n(n+1)(2n+1)}{6} - n$ | M1 A1 | using standard result

$= \frac{n(n+1)(2n+1) - 6n}{6}$ | M1 | combining over common denominator

$= \frac{n[(n+1)(2n+1) - 6]}{6}$ | A1 | factoring $n$

$= \frac{n(2n^2 + 3n - 5)}{6} = \frac{n(2n+5)(n-1)}{6}$ | A1 | fully factorised

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

\hfill \mbox{\textit{OCR MEI FP1 2006 Q3 [6]}}
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