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Find the closed form of \(\displaystyle \displaystyle \sum_{r=1}^{n} \dfrac{1-\cos\left(\frac{(2r-1)\pi}{2n}\right)} {\left[1+\cos\left(\frac{(2r-1)\pi}{2n}\right)\right]\left[5+3\cos\left(\frac{(2r-1)\pi}{2n}\right)\right]^2} \)

\[\large \sum_{r=1}^{n} \dfrac{1-\cos\left(\frac{(2r-1)\pi}{2n}\right)} {\left[1+\cos\left(\frac{(2r-1)\pi}{2n}\right)\right]\left[5+3\cos\left(\frac{(2r-1)\pi}{2n}\right)\right]^2} = \dfrac{n(8n-11)}{16} \]

Prove that for integer \(n>1\), the equation above holds true.

Please try not to use induction (assuming it's possible).


This problem was copied from another math forum but no one responded to it.


This is a part of the set Formidable Series and Integrals.

Note by Aditya Kumar
7 months, 3 weeks ago

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@Aditya Kumar bache ki jaan lo ge kya thoda to hint de the Bhaiya. Lakshya Sinha · 7 months, 3 weeks ago

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have you tried roots of unity Aareyan Manzoor · 7 months, 3 weeks ago

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@Aareyan Manzoor How does that help? We can only use that if we can split out all the expressions and calculate them separately. \( \sum \dfrac f{g \cdot h} \ne \sum f \cdot \sum \dfrac1g \cdot \sum \dfrac1h \). Pi Han Goh · 7 months, 3 weeks ago

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@Pi Han Goh I havent tried this yet but we could use partial fractions and then f'/f or maybe chebyshev polynomials of the first kind..... Aareyan Manzoor · 7 months, 3 weeks ago

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