CAIE FP1 2012 November — Question 4 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeProving standard summation formulae
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring method of differences, algebraic manipulation to derive standard summation formulae, and pattern recognition to sum an alternating series. While the individual techniques are standard for FP1, the final part requires insight to split odd/even terms and apply the derived formulae creatively, making it moderately challenging but still within typical Further Maths scope.
Spec4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series
4 Let \(\mathrm { f } ( r ) = r ( r + 1 ) ( r + 2 )\). Show that $$\mathrm { f } ( r ) - \mathrm { f } ( r - 1 ) = 3 r ( r + 1 )$$ Hence show that \(\sum _ { r = 1 } ^ { n } r ( r + 1 ) = \frac { 1 } { 3 } n ( n + 1 ) ( n + 2 )\). Using the standard result for \(\sum _ { r = 1 } ^ { n } r\), deduce that \(\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )\). Find the sum of the series $$1 ^ { 2 } + 2 \times 2 ^ { 2 } + 3 ^ { 2 } + 2 \times 4 ^ { 2 } + 5 ^ { 2 } + 2 \times 6 ^ { 2 } + \ldots + 2 ( n - 1 ) ^ { 2 } + n ^ { 2 }$$ where \(n\) is odd.
Question 4:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(r(r+1)(r+2)-(r-1)r(r+1) = r(r+1)(r+2-r+1) = 3r(r+1)\) (AG)B1 Verifies given result; part total 1
\(\sum_{r=1}^{n} r(r+1) = \frac{1}{3}\{[f(n)-f(n-1)]+[f(n-1)-f(n-2)]+\ldots+[f(1)-f(0)]\}\)M1 Method of differences
\(= \frac{1}{3}n(n+1)(n+2)\) (AG) — Award B1 if 'not hence'A1 Part total 2
\(\sum_{r=1}^{n} r^2 = \sum_{r=1}^{n} r(r+1) - \sum_{r=1}^{n} r = \frac{n(n+1)(n+2)}{3} - \frac{n(n+1)}{2}\)M1 Subtracts \(\sum r\)
\(= \frac{1}{6}n(n+1)(2n+4-3) = \frac{1}{6}n(n+1)(2n+1)\) (AG)A1 Part total 2
\((1^2+2^2+\ldots+n^2)+(2^2+4^2+\ldots+(n-1)^2) =\)M1 Splits series into two
\(\dfrac{n(n+1)(2n+1)}{6} + \dfrac{4\cdot\frac{n-1}{2}\cdot\frac{n+1}{2}\cdot n}{6} = \ldots = \frac{1}{2}n^2(n+1)\)M1A1 Applies sum of squares formula; part total 3; Total [8] 
## Question 4:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $r(r+1)(r+2)-(r-1)r(r+1) = r(r+1)(r+2-r+1) = 3r(r+1)$ (AG) | B1 | Verifies given result; part total 1 |
| $\sum_{r=1}^{n} r(r+1) = \frac{1}{3}\{[f(n)-f(n-1)]+[f(n-1)-f(n-2)]+\ldots+[f(1)-f(0)]\}$ | M1 | Method of differences |
| $= \frac{1}{3}n(n+1)(n+2)$ (AG) — Award B1 if 'not hence' | A1 | Part total 2 |
| $\sum_{r=1}^{n} r^2 = \sum_{r=1}^{n} r(r+1) - \sum_{r=1}^{n} r = \frac{n(n+1)(n+2)}{3} - \frac{n(n+1)}{2}$ | M1 | Subtracts $\sum r$ |
| $= \frac{1}{6}n(n+1)(2n+4-3) = \frac{1}{6}n(n+1)(2n+1)$ (AG) | A1 | Part total 2 |
| $(1^2+2^2+\ldots+n^2)+(2^2+4^2+\ldots+(n-1)^2) =$ | M1 | Splits series into two |
| $\dfrac{n(n+1)(2n+1)}{6} + \dfrac{4\cdot\frac{n-1}{2}\cdot\frac{n+1}{2}\cdot n}{6} = \ldots = \frac{1}{2}n^2(n+1)$ | M1A1 | Applies sum of squares formula; part total 3; **Total [8]** |

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4 Let $\mathrm { f } ( r ) = r ( r + 1 ) ( r + 2 )$. Show that

$$\mathrm { f } ( r ) - \mathrm { f } ( r - 1 ) = 3 r ( r + 1 )$$

Hence show that $\sum _ { r = 1 } ^ { n } r ( r + 1 ) = \frac { 1 } { 3 } n ( n + 1 ) ( n + 2 )$.

Using the standard result for $\sum _ { r = 1 } ^ { n } r$, deduce that $\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$.

Find the sum of the series

$$1 ^ { 2 } + 2 \times 2 ^ { 2 } + 3 ^ { 2 } + 2 \times 4 ^ { 2 } + 5 ^ { 2 } + 2 \times 6 ^ { 2 } + \ldots + 2 ( n - 1 ) ^ { 2 } + n ^ { 2 }$$

where $n$ is odd.

\hfill \mbox{\textit{CAIE FP1 2012 Q4 [8]}}
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