11. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and for each $n \in \mathbf{N}_{+}, a_{n} 、 a_{n+1}$ are the two roots of the equation $x^{2}+3 n x+b_{n}=0$. Then $\sum_{k=1}^{20} b_{k}=$ $\qquad$
11. 6385.

For each $n \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}=-3 n, \\
a_{n} a_{n+1}=b_{n} .
\end{array}
$$

Rewrite equation (1) as
$$
a_{n+1}+\frac{3(n+1)}{2}-\frac{3}{4}=-\left(a_{n}+\frac{3 n}{2}-\frac{3}{4}\right) \text {. }
$$

Therefore, $\left\{a_{n}+\frac{3 n}{2}-\frac{3}{4}\right\}$ is a geometric sequence with a common ratio of -1.
Hence, $a_{n}+\frac{3 n}{2}-\frac{3}{4}=(-1)^{n-1} \frac{7}{4}$, which means
$$
\begin{array}{l}
a_{n}=-\frac{3(2 n-1)}{4}+(-1)^{n-1} \frac{7}{4}, \\
a_{n+1}=-\frac{3(2 n+1)}{4}+(-1)^{n} \frac{7}{4} .
\end{array}
$$

Then $b_{n}=a_{n} a_{n+1}=\frac{9}{4} n^{2}-\frac{29}{8}+(-1)^{n} \frac{21}{8}$.
Therefore, $\sum_{k=1}^{20} b_{k}=6385$.