# Sum Series using De Moivres Theorem

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#1
Struggling on a question at the moment. Appreciate any help.

I'm on b at the moment. So far I've got
C = [1/2 cos theta - 1/4 - 1/4 cos theta(n+1) + 1/8 cos ntheta ] / (5/4 - cos theta)
Last edited by Toast210; 2 weeks ago
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2 weeks ago
#2
(Original post by Toast210)
Struggling on a question at the moment. Appreciate any help.

I'm on b at the moment. So far I've got
C = [1/2 cos theta - 1/4 - 1/4 cos theta(n+1) + 1/8 cos ntheta ] / (5/4 - cos theta)
Really need to see the full working, but I suspect that when you summed the GP, you used a ratio of instead of , and the 1/2 needs to be raised to the power of n as well in the summation formula - at a guess.
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#3
(Original post by ghostwalker)
Really need to see the full working, but I suspect that when you summed the GP, you used a ratio of instead of , and the 1/2 needs to be raised to the power of n as well in the summation formula - at a guess.
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2 weeks ago
#4
(Original post by Toast210)
As suspected, you've got r correct, but haven't raised it to the power of n corrrectly; should be
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#5
(Original post by ghostwalker)
As suspected, you've got r correct, but haven't raised it to the power of n corrrectly; should be
Could you explain why?
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2 weeks ago
#6
(Original post by Toast210)
Could you explain why?
Because the sum of geo sequence result contains an r^n in there ... so what is (1/2 e^itheta)^n ?
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2 weeks ago
#7
(Original post by Toast210)
Could you explain why?
As part of the formula for a finite sum of the geometric series the common ratio is raised to the power of n. In this case the common ratio is , so raising that to the power of n gives us
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