11. (20 points) Let
$$
\begin{array}{l}
f(x)=a_{1} x^{2013}+a_{2} x^{2012}+\cdots+a_{2013} x+a_{2014} \\
=x^{13}\left(x^{10}+x^{2}+x\right)^{2000}, \\
b_{0}=1, b_{1}=2, b_{n+2}+b_{n}=b_{n+1}(n \in \mathbf{N}) .
\end{array}
$$

Find the value of $\sum_{i=1}^{2013} a_{i} b_{i}$.
11. From the recurrence relation of $\left\{b_{n}\right\}$, we have
$$
\begin{array}{l}
b_{2}=1, b_{3}=-1, b_{4}=-2, b_{5}=-1, \\
b_{6}=1=b_{0}, b_{7}=2=b_{1} .
\end{array}
$$

Thus, $b_{6 k}=1, b_{6 k+1}=2, b_{6 k+2}=1, b_{6 k+3}=-1$, $b_{6 k+4}=-2, b_{6 k+5}=-1(k \in \mathbf{N})$.
Let $\lambda=\frac{1}{2}+\frac{\sqrt{3}}{2}i$. Then
$$
\lambda^{6}=1, \lambda^{10}=-\lambda \text{. }
$$

Therefore, $f(\lambda)=\lambda^{13}\left(\lambda^{10}+\lambda^{2}+\lambda\right)^{200}=\lambda^{13}\left(\lambda^{2}\right)^{200}$
$$
=\lambda^{413}=\frac{1}{2}-\frac{\sqrt{3}}{2}i.
$$

Thus, $\operatorname{Re} f(\lambda)=\frac{1}{2}$.
From $f(\lambda)=\sum_{i=1}^{2014} a_{i} \lambda^{2014-i}$, we know
$\operatorname{Re} f(\lambda)$
$$
=-a_{1}-\frac{1}{2} a_{2}+\frac{1}{2} a_{3}+a_{4}+\frac{1}{2} a_{5}-\frac{1}{2} a_{6} \cdots+\frac{1}{2} a_{2013}+a_{2014}.
$$

Since $a_{2014}=f(0)=0$, we have
$\operatorname{Re} f(\lambda)=-\frac{1}{2} \sum_{i=1}^{2013} a_{i} b_{i}=\frac{1}{2}$.
Thus, $\sum_{i=1}^{2013} a_{i} b_{i}=-1$.