8. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=-1$, and $a_{n}+a_{n+1}+4 n+2=0\left(n \in \mathbf{N}_{+}\right)$, the sum of the first $n$ terms of the subsequence formed by the odd-indexed terms is $\qquad$ .
8. $a_{n}=-2 n^{2}+n$.

From $a_{n}+a_{n+1}+4 n+2=0$, we can get $a_{n-1}+a_{n}+4 n-2=0$. Subtracting these two equations, we have $a_{n+1}-a_{n-1}=-4$, so $\left\{a_{2 n-1}\right\}$ is an arithmetic sequence with the first term $a_{1}$ and common difference -4, $a_{2 n-1}=-4 n+3$. Therefore, the sum of the odd terms equals $\frac{n}{2}\left[a_{1}+a_{2 n-1}\right]=-2 n^{2}+n$.