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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
1 year, 11 months ago

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@Aditya Kumar bache ki jaan lo ge kya thoda to hint de the Bhaiya.

- 1 year, 11 months ago

have you tried roots of unity

- 1 year, 11 months ago

How does that help? We can only use that if we can split out all the expressions and calculate them separately. $$\sum \dfrac f{g \cdot h} \ne \sum f \cdot \sum \dfrac1g \cdot \sum \dfrac1h$$.

- 1 year, 11 months ago

I havent tried this yet but we could use partial fractions and then f'/f or maybe chebyshev polynomials of the first kind.....

- 1 year, 11 months ago