2. Given in $\triangle A B C$, $A B=2, A C=1, \angle B A C=120^{\circ}, O$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{A O}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}$, then $\lambda+\mu=$ $\qquad$ .
2. $\frac{13}{6}$.

Note that $O$ is the circumcenter of $\triangle A B C$, then we have $\overrightarrow{A O} \cdot \overrightarrow{A B}=\frac{1}{2}|\overrightarrow{A B}|^{2}=2$, similarly we get $\overrightarrow{A O} \cdot \overrightarrow{A C}=\frac{1}{2}$. It is also easy to find that $\overrightarrow{A B} \cdot \overrightarrow{A C}=-1$, thus
$$
\left\{\begin{array}{l}
2=\overrightarrow{A O} \cdot \overrightarrow{A B}=\lambda|\overrightarrow{A B}|^{2}+\mu \overrightarrow{A C} \cdot \overrightarrow{A B}=4 \lambda-\mu, \\
\frac{1}{2}=\overrightarrow{A O} \cdot \overrightarrow{A C}=\lambda \overrightarrow{A B} \cdot \overrightarrow{A C}+\mu|\overrightarrow{A C}|^{2}=-\lambda+\mu .
\end{array}\right.
$$

Solving these equations, we get $\lambda=\frac{5}{6}, \mu=\frac{4}{3}$, therefore, $\lambda+\mu=\frac{13}{6}$.