AQA FP1 2006 June — Question 3 4 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2006
SessionJune
Marks4
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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 (∑r² and ∑r) that students memorize for FP1. It requires splitting the sum, applying two formulae, factorizing, and identifying k=1/3. While it's a proof question, it follows a completely routine procedure with no problem-solving or insight needed—easier than average even for Further Maths.
Spec4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series
3 Show that $$\sum _ { r = 1 } ^ { n } \left( r ^ { 2 } - r \right) = k n ( n + 1 ) ( n - 1 )$$ where \(k\) is a rational number.
Question 3:
AnswerMarks Guidance
WorkingMarks Guidance
\(\Sigma(r^2-r) = \Sigma r^2 - \Sigma r\)M1
At least one linear factor foundm1
\(\Sigma(r^2-r) = \frac{1}{6}n(n+1)(2n+1-3)\)m1 OE
\(= \frac{1}{3}n(n+1)(n-1)\)A1 (4) 
## Question 3:
| Working | Marks | Guidance |
|---------|-------|----------|
| $\Sigma(r^2-r) = \Sigma r^2 - \Sigma r$ | M1 | |
| At least one linear factor found | m1 | |
| $\Sigma(r^2-r) = \frac{1}{6}n(n+1)(2n+1-3)$ | m1 | OE |
| $= \frac{1}{3}n(n+1)(n-1)$ | A1 (4) | |

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3 Show that

$$\sum _ { r = 1 } ^ { n } \left( r ^ { 2 } - r \right) = k n ( n + 1 ) ( n - 1 )$$

where $k$ is a rational number.

\hfill \mbox{\textit{AQA FP1 2006 Q3 [4]}}
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