2. Let the sequence $\left\{a_{n}\right\}$ satisfy, $a_{1}=1, a_{n+1}=5 a_{n}+1(n=1,2, \cdots)$, then $\sum_{n=1}^{2018} a_{n}=$
Calculate $\frac{5^{2019}}{16}-\frac{8077}{16}$.
Solution From $a_{n+1}=5 a_{n}+1 \Rightarrow a_{n+1}+\frac{1}{4}=5\left(a_{n}+\frac{1}{4}\right) \Rightarrow a_{n}=\frac{5^{n}}{4}-\frac{1}{4}$, so
$$
\sum_{n=1}^{2018} a_{n}=\frac{1}{4}\left(5^{1}+5^{2}+\cdots+5^{2018}\right)-\frac{2018}{4}=\frac{5}{16}\left(5^{2018}-1\right)-\frac{2018}{4}=\frac{5^{2019}}{16}-\frac{8077}{16}
$$