5. Let the first two terms of the sequence $\left\{a_{n}\right\}$ be $a_{1}=2$, $a_{2}=4$. Let $b_{n}=a_{n+1}-a_{n}$. If the sequence $\left\{b_{n}\right\}$ is an arithmetic sequence with a common difference of 1, then $a_{2020}=$ $\qquad$
5.2041211.

It is known that $b_{1}=a_{2}-a_{1}=2$.
From the common difference of $\left\{b_{n}\right\}$ being 1, we know that
$$
b_{n}=n+1 \text {. }
$$
$$
\begin{array}{l}
\text { Hence } a_{2020}=a_{1}+\sum_{i=1}^{2019} b_{i} \\
=2+2+3+\cdots+2020 \\
=2041211 \text {. }
\end{array}
$$