Let $n$ be a positive integer and $\zeta$ be a primitive $n$th root of unity. It is not hard to show that \begin{align*} \sum_{k=1}^{n-1}\frac{\zeta^k}{1-\zeta^k}=\frac{1-n}{2}. \end{align*} Since $\zeta^n=1$, we have \begin{align*} \frac{\zeta^k}{1-\zeta^k}+\frac{\zeta^{n-k}}{1-\zeta^{n-k}} =\frac{\zeta^k}{1-\zeta^k}+\frac{\zeta^{-k}}{1-\zeta^{-k}} =-1, \end{align*} and so \begin{align*} \sum_{k=1}^{n-1}\frac{\zeta^k}{1-\zeta^k}=\frac{1}{2} \sum_{k=1}^{n-1}\left(\frac{\zeta^k}{1-\zeta^k}+\frac{\zeta^{-k}}{1-\zeta^{-k}}\right) =\frac{1-n}{2}. \end{align*} Let $\omega$ be a primitive $(3n+2)$th root of unity. By the same method, we can also get \begin{align*} \sum_{k=0}^{2n+1}(-1)^k\omega^{\frac{k(3k+1)}{2}}= \sum_{k=0}^{n}\left((-1)^k\omega^{\frac{k(3k+1)}{2}}+(-1)^{2n+1-k} \omega^{\frac{(2n+1-k)(3(2n+1-k)+1)}{2}}\right)=0, \end{align*} because \begin{align*} \omega^{\frac{(2n+1-k)(3(2n+1-k)+1)}{2}}=\omega^{\frac{k(3k+1)}{2}+(3n+2)(2n-2k+1)}= \omega^{\frac{k(3k+1)}{2}}. \end{align*} Note that the above two sums possess natural symmetries.

**Question.**
Let $\omega=e^\frac{2\pi i}{3n+2}$ be the primitive $(3n+2)$th root of unity.
Numerical calculation suggests that
\begin{align*}
\sum_{k=1}^{2n+1}\frac{(-1)^k\omega^{\frac{k(3k+1)}{2}}}{1-\omega^{3k}}=-\frac{n+1}{2}.
\end{align*}
Is this identity true? If so, how to prove it?
Unfortunately, this sum loses a natural symmetry.
Hints, references or proofs are all welcome!

**Comments:**
Nemo proved that for arbitrary $w$,
\begin{align*}
\sum _{k=1}^{2 n+1} \frac{(-1)^k w^{k (3 k+1)/2}}{1-w^{3 k}}&=-\sum _{k=0}^{2 n} \frac{(-1)^k w^{\frac{1}{2} (k+2) (3 k+1)}}{1-w^{3 k+1}}\\
&+\frac{1}{2} \sum _{k=0}^{2 n} \left(\frac{(-1)^k w^{(3 k+1) (n+1)}}{1-w^{(3k+1)/2}}+\frac{w^{(3 k+1) (n+1)}}{w^{(3k+1)/2}+1}\right)\\
&+\frac{1}{2} \sum _{k=1}^{2 n+1} \left(\frac{(-1)^k w^{k/2}}{1-w^{3k/2}}+\frac{w^{k/2}}{w^{3k/2}+1}\right).
\end{align*}
This identity has been reproved by GH from MO below.

Letting $k\to 2n+1-k$ in the following sum gives \begin{align*} \sum _{k=0}^{2 n} \frac{(-1)^k w^{\frac{1}{2} (k+2) (3 k+1)}}{1-w^{3 k+1}} =\sum _{k=1}^{2 n+1} \frac{(-1)^k w^{k (3 k+1)/2}}{1-w^{3 k}}, \end{align*} where we have used the fact $w^{3n+2}=1$. Also, we have (letting $k\to 2n+1-k$) \begin{align*} \sum _{k=0}^{2 n} \left(\frac{(-1)^k w^{(3 k+1) (n+1)}}{1-w^{(3k+1)/2}}+\frac{w^{(3 k+1) (n+1)}}{w^{(3k+1)/2}+1}\right)= \sum _{k=1}^{2 n+1} \left(\frac{(-1)^k w^{k/2}}{1-w^{3k/2}}+\frac{w^{k/2}}{w^{3k/2}+1}\right). \end{align*} So it suffices to show that \begin{align*} \sum _{k=1}^{2 n+1} \left(\frac{(-1)^k w^{k/2}}{1-w^{3k/2}}+\frac{w^{k/2}}{w^{3k/2}+1}\right)=-n-1, \end{align*} which was proved by Fedor Petrov.