11 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 $\sum_{k=1}^{20} b_{k}=$ $\qquad$.
For each $n \in \mathbf{N}^{*}$,
$$
a_{n}+a_{n+1}=-3 n, a_{n} a_{n+1}=b_{n},
$$

we can see that
$$
a_{n+1}+\frac{3(n+1)}{2}-\frac{3}{4}=-\left(a_{n}+\frac{3 n}{2}-\frac{3}{4}\right),
$$

thus $\left\{a_{n}+\frac{3 n}{2}-\frac{3}{4}\right\}$ is a geometric sequence with a common ratio of -1, hence
$$
a_{n}+\frac{3 n}{2}-\frac{3}{4}=(-1)^{n-1} \frac{7}{4},
$$

which means
$$
a_{n}=-\frac{3(2 n-1)}{4}+(-1)^{n-1} \cdot \frac{7}{4},
$$

then
$$
a_{n+1}=-\frac{3(2 n+1)}{4}+(-1)^{n} \cdot \frac{7}{4},
$$

thus
$$
b_{n}=a_{n} a_{n+1}=\frac{9}{4} n^{2}-\frac{29}{8}+(-1)^{n} \cdot \frac{21}{8},
$$

therefore $\sum_{k=1}^{20} b_{k}=6385$.