5 State the sum of the series \(z + z ^ { 2 } + z ^ { 3 } + \ldots + z ^ { n }\), for \(z \neq 1\).
By letting \(z = \cos \theta + \mathrm { i } \sin \theta\), show that
$$\cos \theta + \cos 2 \theta + \cos 3 \theta + \ldots + \cos n \theta = \frac { \sin \frac { 1 } { 2 } n \theta } { \sin \frac { 1 } { 2 } \theta } \cos \frac { 1 } { 2 } ( n + 1 ) \theta$$
where \(\sin \frac { 1 } { 2 } \theta \neq 0\).
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Question 5:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\frac{z(z^n-1)}{z-1}\) (OE) B1
[1]
\(\cos\theta+\cos 2\theta+\cdots+\cos n\theta=\text{Re}\left\{\frac{z(z^n-1)}{(z-1)}\right\}\) M1
\(=\text{Re}\left\{\frac{z^{\frac{1}{2}}(z^n-1)}{\left(z^{\frac{1}{2}}-z^{-\frac{1}{2}}\right)}\right\}\) M1
\(=\text{Re}\left\{\frac{z^{n+\frac{1}{2}}-z^{\frac{1}{2}}}{2i\sin\frac{1}{2}\theta}\right\}\) A1
\(=\frac{\sin\left(n+\frac{1}{2}\right)\theta-\sin\frac{1}{2}\theta}{2\sin\frac{1}{2}\theta}\) A1
\(=\frac{2\cos\frac{1}{2}(n+1)\theta\sin\frac{1}{2}n\theta}{2\sin\frac{1}{2}\theta}\) M1A1
Both previous M marks required
\(=\frac{\cos\frac{1}{2}(n+1)\theta\sin\frac{1}{2}n\theta}{\sin\frac{1}{2}\theta}\) A1
[7] AG
Question 5 (Alternative i):
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(\Sigma = \text{Re}\left\{\frac{z - z^{n+1}}{1-z}\right\}\) M1
\(= \text{Re}\left\{\frac{e^{i\theta} - e^{i(n+1)\theta}}{1 - e^{i\theta}} \times \frac{1 - e^{-i\theta}}{1 - e^{-i\theta}}\right\}\) M1
\(= \frac{\cos\theta - \cos(n+1)\theta - 1 + \cos n\theta}{2 - 2\cos\theta}\) A1A1
Numerator and denominator
\(= \frac{-2\sin^2\left(\frac{\theta}{2}\right) + 2\sin\left(\frac{2n+1}{2}\right)\theta\sin\left(\frac{\theta}{2}\right)}{4\sin^2\left(\frac{\theta}{2}\right)}\) A1
\(= \frac{1}{2\sin\left(\frac{\theta}{2}\right)}\left\{\sin\left(n+\frac{1}{2}\right)\theta - \sin\left(\frac{\theta}{2}\right)\right\}\) M1
\(= \frac{\cos\left(\frac{n+1}{2}\right)\theta\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\) A1
CAO, AG
Total [7]
Question 5 (Alternative ii):
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(\Sigma = \text{Re}\left\{\frac{z^{n+1} - z}{z - 1}\right\}\) M1
\(= \text{Re}\left\{\frac{(\cos\theta + i\sin\theta) - (\cos[n+1]\theta + i\sin[n+1]\theta)}{(1-\cos\theta) - i\sin\theta} \times \frac{(1-\cos\theta)+i\sin\theta}{(1-\cos\theta)+i\sin\theta}\right\}\) M1
Numerator and denominator correctly simplified A1A1
\(= \frac{\sin\left(\frac{n\theta}{2}\right)}{2\sin^2\left(\frac{\theta}{2}\right)}\left\{\sin\left[\frac{n+2}{2}\right]\theta(1-\cos\theta) + \cos\left[\frac{n+2}{2}\right]\theta\sin\theta\right\}\) A1
\(= \frac{\sin\left(\frac{n\theta}{2}\right)}{2\sin^2\left(\frac{\theta}{2}\right)}\left\{\sin\left[\frac{n+2}{2}\right]\theta - \sin\left(\frac{n\theta}{2}\right)\right\}\) M1
\(= \frac{\cos\left(\frac{n+1}{2}\right)\theta\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\) A1
CAO, AG
Total [7]
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## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{z(z^n-1)}{z-1}$ (OE) | B1 | [1] |
| $\cos\theta+\cos 2\theta+\cdots+\cos n\theta=\text{Re}\left\{\frac{z(z^n-1)}{(z-1)}\right\}$ | M1 | |
| $=\text{Re}\left\{\frac{z^{\frac{1}{2}}(z^n-1)}{\left(z^{\frac{1}{2}}-z^{-\frac{1}{2}}\right)}\right\}$ | M1 | |
| $=\text{Re}\left\{\frac{z^{n+\frac{1}{2}}-z^{\frac{1}{2}}}{2i\sin\frac{1}{2}\theta}\right\}$ | A1 | |
| $=\frac{\sin\left(n+\frac{1}{2}\right)\theta-\sin\frac{1}{2}\theta}{2\sin\frac{1}{2}\theta}$ | A1 | |
| $=\frac{2\cos\frac{1}{2}(n+1)\theta\sin\frac{1}{2}n\theta}{2\sin\frac{1}{2}\theta}$ | M1A1 | Both previous M marks required |
| $=\frac{\cos\frac{1}{2}(n+1)\theta\sin\frac{1}{2}n\theta}{\sin\frac{1}{2}\theta}$ | A1 | [7] AG |
# Question 5 (Alternative i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\Sigma = \text{Re}\left\{\frac{z - z^{n+1}}{1-z}\right\}$ | M1 | |
| $= \text{Re}\left\{\frac{e^{i\theta} - e^{i(n+1)\theta}}{1 - e^{i\theta}} \times \frac{1 - e^{-i\theta}}{1 - e^{-i\theta}}\right\}$ | M1 | |
| $= \frac{\cos\theta - \cos(n+1)\theta - 1 + \cos n\theta}{2 - 2\cos\theta}$ | A1A1 | Numerator and denominator |
| $= \frac{-2\sin^2\left(\frac{\theta}{2}\right) + 2\sin\left(\frac{2n+1}{2}\right)\theta\sin\left(\frac{\theta}{2}\right)}{4\sin^2\left(\frac{\theta}{2}\right)}$ | A1 | |
| $= \frac{1}{2\sin\left(\frac{\theta}{2}\right)}\left\{\sin\left(n+\frac{1}{2}\right)\theta - \sin\left(\frac{\theta}{2}\right)\right\}$ | M1 | |
| $= \frac{\cos\left(\frac{n+1}{2}\right)\theta\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}$ | A1 | CAO, AG |
| Total | [7] | |
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# Question 5 (Alternative ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\Sigma = \text{Re}\left\{\frac{z^{n+1} - z}{z - 1}\right\}$ | M1 | |
| $= \text{Re}\left\{\frac{(\cos\theta + i\sin\theta) - (\cos[n+1]\theta + i\sin[n+1]\theta)}{(1-\cos\theta) - i\sin\theta} \times \frac{(1-\cos\theta)+i\sin\theta}{(1-\cos\theta)+i\sin\theta}\right\}$ | M1 | |
| Numerator and denominator correctly simplified | A1A1 | |
| $= \frac{\sin\left(\frac{n\theta}{2}\right)}{2\sin^2\left(\frac{\theta}{2}\right)}\left\{\sin\left[\frac{n+2}{2}\right]\theta(1-\cos\theta) + \cos\left[\frac{n+2}{2}\right]\theta\sin\theta\right\}$ | A1 | |
| $= \frac{\sin\left(\frac{n\theta}{2}\right)}{2\sin^2\left(\frac{\theta}{2}\right)}\left\{\sin\left[\frac{n+2}{2}\right]\theta - \sin\left(\frac{n\theta}{2}\right)\right\}$ | M1 | |
| $= \frac{\cos\left(\frac{n+1}{2}\right)\theta\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}$ | A1 | CAO, AG |
| Total | [7] | |
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5 State the sum of the series $z + z ^ { 2 } + z ^ { 3 } + \ldots + z ^ { n }$, for $z \neq 1$.
By letting $z = \cos \theta + \mathrm { i } \sin \theta$, show that
$$\cos \theta + \cos 2 \theta + \cos 3 \theta + \ldots + \cos n \theta = \frac { \sin \frac { 1 } { 2 } n \theta } { \sin \frac { 1 } { 2 } \theta } \cos \frac { 1 } { 2 } ( n + 1 ) \theta$$
where $\sin \frac { 1 } { 2 } \theta \neq 0$.
\hfill \mbox{\textit{CAIE FP1 2014 Q5 [8]}}