OCR FP1 2008 June — Question 5 6 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionJune
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 of r²(r-1) = r³ - r², then applying the known formulae for Σr³ and Σr². The algebraic manipulation and factorisation at the end is routine. While it's a Further Maths question, it's a standard textbook exercise with no novel insight required, making it slightly easier than average overall.
Spec4.06a Summation formulae: sum of r, r^2, r^3

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

AnswerMarks Guidance
\(\frac{1}{4}n^2(n+1)^2 - \frac{1}{6}n(n+1)(2n+1)\)M1 Express as difference of two series
M1Use standard results
A1Correct unsimplified answer
\(\frac{1}{12}n(n+1)(3n+2)(n-1)\)M1 Attempt to factorise
A1At least factor of \(n(n+1)\)
A1Obtain correct answer
6 marks
$\frac{1}{4}n^2(n+1)^2 - \frac{1}{6}n(n+1)(2n+1)$ | M1 | Express as difference of two series
| M1 | Use standard results
| A1 | Correct unsimplified answer

$\frac{1}{12}n(n+1)(3n+2)(n-1)$ | M1 | Attempt to factorise
| A1 | At least factor of $n(n+1)$
| A1 | Obtain correct answer
| **6 marks**
5 Find $\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r - 1 )$, expressing your answer in a fully factorised form.

\hfill \mbox{\textit{OCR FP1 2008 Q5 [6]}}