2. Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
\begin{array}{l}
a_{1}=1, a_{n+1}=5 a_{n}+1(n=1,2, \cdots) \\
\text { then } \sum_{n=1}^{2018} a_{n}=
\end{array}
$$
$\begin{array}{l} \text { 2. } \frac{5^{2019}}{16}-\frac{8077}{16} . \\ \text { Given } a_{n+1}=5 a_{n}+1 \\ \Rightarrow a_{n+1}+\frac{1}{4}=5\left(a_{n}+\frac{1}{4}\right) \\ \Rightarrow a_{n}=\frac{5^{n}}{4}-\frac{1}{4} . \\ \text { Then } \sum_{n=1}^{2018} a_{n}=\frac{1}{4} \sum_{n=1}^{2018} 5^{n}-\frac{2018}{4} \\ = \frac{5^{2019}}{16}-\frac{8077}{16} .\end{array}$