CAIE FP1 2010 June — Question 2 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicSequences and Series
TypePartial Fractions and Telescoping Series
DifficultyChallenging +1.8 This is a Further Maths question requiring recognition of a telescoping series structure from a given trigonometric identity, algebraic manipulation to derive a finite sum formula, and application to determine convergence. While the identity is provided (reducing discovery difficulty), students must manipulate it correctly, recognize the telescoping pattern, and apply the result to a specific case. The multi-step reasoning and proof elements place it well above average difficulty, though the scaffolding prevents it from reaching the highest difficulty levels.
Spec4.06b Method of differences: telescoping series
2 By considering the identity $$\cos [ ( 2 n - 1 ) \alpha ] - \cos [ ( 2 n + 1 ) \alpha ] \equiv 2 \sin \alpha \sin 2 n \alpha$$ show that if \(\alpha\) is not an integer multiple of \(\pi\) then $$\sum _ { n = 1 } ^ { N } \sin ( 2 n \alpha ) = \frac { 1 } { 2 } \cot \alpha - \frac { 1 } { 2 } \operatorname { cosec } \alpha \cos [ ( 2 N + 1 ) \alpha ]$$ Deduce that the infinite series $$\sum _ { n = 1 } ^ { \infty } \sin \left( \frac { 2 } { 3 } n \pi \right)$$ does not converge.
AnswerMarks Guidance
\(2 \sin \alpha \sum_{n=1}^{N} \sin(2n\alpha) = \cos \alpha - \cos[(2N + 1)\alpha]\)M1A1
\(\Rightarrow\) displayed result (AG)M1A1 [4]
\(\cos(2N + 1)\pi/3\) oscillates finitely as \(n \to \infty\) ⟹ \(\sum_{n=1}^{\infty} \sin(2n\pi/3)\) does not converge (CWO)B1
Require \(\alpha = \frac{\pi}{3}\), 'oscillate' or values of \(\cos(2N + 1)\frac{\pi}{3}\) given as \(\frac{1}{2}\) or \(-1\)[1] 
$2 \sin \alpha \sum_{n=1}^{N} \sin(2n\alpha) = \cos \alpha - \cos[(2N + 1)\alpha]$ | M1A1 |

$\Rightarrow$ displayed result (AG) | M1A1 | [4]

$\cos(2N + 1)\pi/3$ oscillates finitely as $n \to \infty$ ⟹ $\sum_{n=1}^{\infty} \sin(2n\pi/3)$ does not converge (CWO) | B1 |

Require $\alpha = \frac{\pi}{3}$, 'oscillate' or values of $\cos(2N + 1)\frac{\pi}{3}$ given as $\frac{1}{2}$ or $-1$ | [1] |

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2 By considering the identity

$$\cos [ ( 2 n - 1 ) \alpha ] - \cos [ ( 2 n + 1 ) \alpha ] \equiv 2 \sin \alpha \sin 2 n \alpha$$

show that if $\alpha$ is not an integer multiple of $\pi$ then

$$\sum _ { n = 1 } ^ { N } \sin ( 2 n \alpha ) = \frac { 1 } { 2 } \cot \alpha - \frac { 1 } { 2 } \operatorname { cosec } \alpha \cos [ ( 2 N + 1 ) \alpha ]$$

Deduce that the infinite series

$$\sum _ { n = 1 } ^ { \infty } \sin \left( \frac { 2 } { 3 } n \pi \right)$$

does not converge.

\hfill \mbox{\textit{CAIE FP1 2010 Q2 [5]}}
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