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I came across this question from an old past paper where no solutions were provided; I only need guidance and I will give it a go...

Given that $w_n=3^{(-n)} \cos⁡2nθ$ [(for n=1,2,3….〗Use De Moivre’s theorem to show that $$1 + w_{1} + w_2 + \cdots + w_{(n-1)} = \frac{(9 - 3 \cos2\theta + 3^{(-n-1)} \cos 2⁡(n-1)\theta-3^{(-n-1)} \cos2n\theta)}{(10 - 6 \cos 2\theta)}$$

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Hint: $w_n$ is the real part of the complex number $z_n = (3)^{-n} (\cos(2n \theta) + i \sin(2n \theta))$. So $1 + w_1 \ldots + w_{n-1}$ is the real part of $1 + z_1 + \ldots + z_{n-1}$.

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  • $\begingroup$ A great hint, I should be there soon, thanks. $\endgroup$ Mar 6 '19 at 9:34
  • $\begingroup$ could there be a binomial expansion part in this? $\endgroup$ Mar 6 '19 at 18:36
  • $\begingroup$ I don't think there is an obvious way to drag the binomial expansion into this. Try to think of another famous identity after having applied De Moivre's theorem. $\endgroup$
    – user649035
    Mar 6 '19 at 22:10
  • $\begingroup$ Is this the key ? Re (z + 1/z) = 2cos (x) $\endgroup$ Mar 6 '19 at 22:18
  • $\begingroup$ It wasn't what I had in mind but it may help. If you let $a = \frac{1}{3} (\cos(2\theta) + i \sin(2\theta))$, how does $1 + z_1 + \ldots + z_{n-1}$ look ? $\endgroup$
    – user649035
    Mar 6 '19 at 22:24

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