14.1. $[9.1$ (15 points), 10.1 (15 points)] Calculate

$$
\sqrt{1+2 \sqrt{1+3 \sqrt{1+\ldots+2017 \sqrt{1+2018 \cdot 2020}}}}
$$

Answer: 3.
Solution. Since $2018 \cdot 2020=2019^{2}-1, 2017 \cdot 2019=2018^{2}-1, \ldots$, we get

$$
\begin{gathered}
\sqrt{1+2 \sqrt{1+3 \sqrt{1+\ldots+2017 \sqrt{1+2018 \cdot 2020}}}}= \\
=\sqrt{1+2 \sqrt{1+3 \sqrt{1+\ldots+2016 \sqrt{1+2017 \cdot 2019}}}}=\ldots=\sqrt{1+2 \cdot 4}=3
\end{gathered}
$$