How can you prove that:∑k=1n−1tan2kπ2n=(n−1)(2n−1)3for every integer n≥1.

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Manraj Horton · Accepted Answer

By a well know formula we have(cos⁡kπ2n+isin⁡kπ2n)2n=(−1)kHence by Binomial theorem we have ([x] is not an integer part, brackets are added for clarity)∑t=02n(2nt)[cos⁡kπ2n]t⋅[isin⁡kπ2n]2n−t=(−1)kNow we consider only imaginary part of this:∑r=0n−1(2n2r+1)[cos⁡kπ2n]2r+1⋅[isin⁡kπ2n]2n−2r−1=0Divide it by [cos⁡kπ2n]2n:∑r=0n−1(2n2r+1)[itan⁡kπ2n]2n−2r−1=0Now multiply by itan⁡kπ2n:∑r=0n−1(2n2r+1)[itan⁡kπ2n]2n−2r=0So [tan⁡kπ2n]2 are roots of the following polynomial:∑r=0n−1(2n2r+1)[−x]n−r=0Hence by Vieta&#039;s formulas sum of it roots is equal to(2n3)(2n1)=(2n−1)(n−1)3

Cicolinif73 · Accepted Answer

I am doing a similar thing for cot⁡ i hope you can reciprocate it.We have(cot⁡(θ)−i)n=cos⁡nθ−sin⁡nθsinn(θ)Equating the real and imaginary parts on both sides we havesin⁡(nθ)sin⁡θ=∑s(n2s+1)(−1)scotn−2s−1(θ)Now take 2n+1 instead of n we have,sin⁡(2n+1)θ=sin2n+1(θ)Pncot2(θ)for 0&amp;lt;θ&amp;lt;π2 and where Pn is the polynomial given by(2n+11)Tn−(2n+13)Tn−1+…Noting that the zeros of Pn are precisely rπ2n+1, n=1,2,…, we have the first identity from the sum of the roots formula.From the inequality sin⁡x&amp;lt;x&amp;lt;tan⁡x we havecot2m(x)&amp;lt;1x2m&amp;lt;(1+cot2(x))mwe have∑r=1ncot2mrπ2n+1&amp;lt;(2n+1)2mπ2m∑r=1n1r2m&amp;lt;∑r=1n(1+cot2rπ2n+1)mTherefore,∑r=1n(1+cot2rπ2n+1)m=∑r=1ncot2mrπ2n+1+O(n2m−1)In other words to find c2m where

How can you prove that: \sum_{k=1}^{n-1}\tan^2\frac{k\pi}{2n}=\frac{(n-1)(2n-1)}{3} for every integer n\geq1.

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