CAIE FP1 2012 November — Question 1 4 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyStandard +0.8 This requires manipulating the standard sum formula for r² by splitting it into two ranges (1 to 2n minus 1 to n), then algebraically simplifying to reach the given form. While the technique is standard for Further Maths, the algebraic manipulation is non-trivial and requires careful factorization, making it moderately challenging but within expected FM scope.
Spec4.06a Summation formulae: sum of r, r^2, r^3
1 Show that \(\sum _ { r = n + 1 } ^ { 2 n } r ^ { 2 } = \frac { 1 } { 6 } n ( 2 n + 1 ) ( 7 n + 1 )\).
Question 1:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\sum_{n+1}^{2n} = \sum_{1}^{2n} - \sum_{1}^{n}\)M1 Use of splitting sum
\(\sum_{1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\)M1 Use of standard formula
\(\frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6}\)A1 Obtains result
\(= \frac{1}{6}n(2n+1)(8n+2-n-1) = \frac{1}{6}n(2n+1)(7n+1)\)A1 AG 
## Question 1:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\sum_{n+1}^{2n} = \sum_{1}^{2n} - \sum_{1}^{n}$ | M1 | Use of splitting sum |
| $\sum_{1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$ | M1 | Use of standard formula |
| $\frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6}$ | A1 | Obtains result |
| $= \frac{1}{6}n(2n+1)(8n+2-n-1) = \frac{1}{6}n(2n+1)(7n+1)$ | A1 | AG |

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1 Show that $\sum _ { r = n + 1 } ^ { 2 n } r ^ { 2 } = \frac { 1 } { 6 } n ( 2 n + 1 ) ( 7 n + 1 )$.

\hfill \mbox{\textit{CAIE FP1 2012 Q1 [4]}}
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Q1 4 Q2 4 Q3 5 Q4 6 Q5 6 Q6 7 Q7 8 Q8 9 Q9 12 Q10 12 Q11 13 Q12 EITHER Q12 OR