2. Given that $S_{n}$ and $T_{n}$ are the sums of the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively, and
$$
\frac{S_{n}}{T_{n}}=\frac{2 n+1}{4 n-2}(n=1,2, \cdots) \text {. }
$$

then $\frac{a_{10}}{b_{3}+b_{18}}+\frac{a_{11}}{b_{6}+b_{15}}=$ $\qquad$
2. $\frac{41}{78}$.

Since $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both arithmetic sequences, we have,
$$
\begin{array}{l}
\frac{a_{10}}{b_{3}+b_{18}}+\frac{a_{11}}{b_{6}+b_{15}}=\frac{a_{10}}{b_{1}+b_{20}}+\frac{a_{11}}{b_{1}+b_{20}} \\
=\frac{a_{1}+a_{20}}{b_{1}+b_{20}}=\frac{S_{20}}{T_{20}}=\frac{2 \times 20+1}{4 \times 20-2}=\frac{41}{78} .
\end{array}
$$