OCR MEI FP1 2005 June — Question 7 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2005
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicSequences and Series
TypeSum of Powers Using Standard Formulae
DifficultyModerate -0.5 This is a straightforward application of standard summation formulae requiring expansion of 3r(r-1) into 3r² - 3r, then applying the formulae for Σr² and Σr. The algebraic manipulation and factorisation are routine for Further Maths students, making it slightly easier than average but still requiring correct technique.
Spec4.06b Method of differences: telescoping series
7 Find \(\sum _ { r = 1 } ^ { n } 3 r ( r - 1 )\), expressing your answer in a fully factorised form.
Question 7:
AnswerMarks Guidance
\(3\sum r^2 - 3\sum r\)M1, A1 Separate sums
\(= 3 \times \frac{1}{6}n(n+1)(2n+1) - 3 \times \frac{1}{2}n(n+1)\)M1, A1 Use of formulae
\(= \frac{1}{2}n(n+1)\left[(2n+1) - 3\right]\)M1 Attempt to factorise, only if earlier M marks awarded
\(= \frac{1}{2}n(n+1)(2n-2)\)— —
\(= n(n+1)(n-1)\)A1 (c.a.o.) [6] Must be fully factorised 
## Question 7:

$3\sum r^2 - 3\sum r$ | M1, A1 | Separate sums

$= 3 \times \frac{1}{6}n(n+1)(2n+1) - 3 \times \frac{1}{2}n(n+1)$ | M1, A1 | Use of formulae

$= \frac{1}{2}n(n+1)\left[(2n+1) - 3\right]$ | M1 | Attempt to factorise, only if earlier M marks awarded

$= \frac{1}{2}n(n+1)(2n-2)$ | — | —

$= n(n+1)(n-1)$ | A1 (c.a.o.) **[6]** | Must be fully factorised

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

\hfill \mbox{\textit{OCR MEI FP1 2005 Q7 [6]}}
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