6. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1$, and for each $n \in \mathbf{N}^{*}, a_{n}, a_{n+1}$ are the roots of the equation $x^{2}+3 n x+b_{n}=0$, then $\sum_{k=1}^{20} b_{k}=$ $\qquad$ .
6. 6385 Detailed Explanation: For each $n \in \mathbf{N}^{*}, a_{n}+a_{n+1}=-3 n$ (1), $a_{n} a_{n+1}=b_{n}$ (2),

Rewrite (1) as $a_{n+1}+\frac{3(n+1)}{2}-\frac{3}{4}=-\left(a_{n}+\frac{3 n}{2}-\frac{3}{4}\right)$, thus $\left\{a_{n}+\frac{3 n}{2}-\frac{3}{4}\right\}$ is a geometric sequence with a common ratio of -1, hence $a_{n}+\frac{3 n}{2}-\frac{3}{4}=(-1)^{n-1} \frac{7}{4}$, i.e., $a_{n}=-\frac{3(2 n-1)}{4}+(-1)^{n-1} \cdot \frac{7}{4}$, $a_{n+1}=-\frac{3(2 n+1)}{4}+(-1)^{n} \cdot \frac{7}{4}$, thus $b_{n}=a_{n} a_{n+1}=\frac{9}{4} n^{2}-\frac{29}{8}+(-1)^{n} \cdot \frac{21}{8} ; \sum_{k=1}^{20} b_{k}=6385$.