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Let $n$ be a positive integer. Evaluate in closed form: $$\begin{align} S_n &= \sum_{k=0}^{n-1}\tan^2{\left( \frac {2k+1}{2n+1} \cdot \frac{\pi}{4} \right)} &+\cot^2{\left( \frac {2k+1}{2n+1} \cdot \frac{\pi}{4} \right)} \end{align}$$

This is an intresting problem which has had me stumped for days. I don't really see complex numbers here so I tried to simplify it using some trig identities and some algebra. $$S_n=\sum_{k=0}^{n-1} \frac{4}{\cos^2\left(\frac{n-k}{2n+1}\cdot \pi\right)}-2$$ After this I tried partial fraction but that feels wrong. Even after further manipulation and pairing terms nothing worthwhile is coming up. The solution key says the answer is $8n^2+6n.$ (although we would not be given the closed form) I also tried induction on it but that didn't work out.

I solved this problem now, but was wondering if there are alternate ways to do it!

Thank you in advance!

Aditya Teraiya
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    See for similar identities and how they are proved. For example, we have $$\sum_{k=1}^n \cot^2 \left( \frac {2k-1}{2n} \frac{\pi}{2} \right) = 2n^2 – n. $$ See this post for a proof. – Dietrich Burde Feb 17 '25 at 19:13
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    $$\cot^2y+\tan^2y=(\cot y-\tan y)^2+2=4\cot^22y+2$$ Now can you utilize https://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series or https://math.stackexchange.com/questions/1562037/question-regarding-fn-cot2-left-frac-pi-n-right-cot2-left-frac2-pin – lab bhattacharjee Feb 17 '25 at 19:15

1 Answers1

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Thank you for the help in comments

By the formula suggested by @labbhattacharjee we have $$S_n=\sum_{k=0}^{n-1}4\cot^2\left(\frac{2k+1}{2n+1}\cdot \frac{\pi}{2}\right)+2=4\cdot \sum_{k=0}^{n-1}\left(\frac{1}{\cot^2\left(\frac{(n-k) \pi}{2n+1}\right)}\right)+2n$$

However from this post (suggested by @DietrichBurde) consider the polynomial $P(x)=\sum_{k=0}^n(-1)^k\binom{2n+1}{2k+1}x^{n-k}\,$ The roots of this polynomial are $$\cot^2\frac{\pi}{2n+1},\quad\cot^2\frac{2\pi}{2n+1},\quad\ldots,\quad\cot^2\frac{n\pi}{2n+1}$$ Which is great because they are simply the reciprocal of what we require. We require $\sum \frac{1}{\alpha_i}, \alpha_i =\cot^2 \frac{i \pi}{2n+1} 1\leq i \leq n$

By Vieta's, we get $\frac{(-1)^{n-1+1}\binom{2n+1}{2n-1}}{(-1)^n}=2n^2+n$

So we get $S_n=8n^2+6n.$

Aditya Teraiya
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