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Part of 2018 PUMaC Algebra A

Problems(1)

2018 PUMaC Algebra A8

Source:

11/25/2018
pq=∑n=1∞12n+6(10−4cos⁡2(πn24))(1−(−1)n)−3cos⁡(πn24)(1+(−1)n)25−16cos⁡2(πn24)\frac{p}{q} = \sum_{n = 1}^\infty \frac{1}{2^{n + 6}} \frac{(10 - 4\cos^2(\frac{\pi n}{24})) (1 - (-1)^n) - 3\cos(\frac{\pi n}{24}) (1 + (-1)^n)}{25 - 16\cos^2(\frac{\pi n}{24})} where pp and qq are relatively prime positive integers. Find p+qp + q.
algebra