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.

#Calculus

Easy Math Editor

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

Department 8 - 5 years, 4 months ago

have you tried roots of unity

Aareyan Manzoor - 5 years, 4 months ago

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

Pi Han Goh - 5 years, 4 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.....

Aareyan Manzoor - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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

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