2. Find the value of the fraction

$$
\frac{2 \cdot 2020}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+2020}}
$$
Solution. Let's denote the denominator of the fraction by $q$. By repeatedly applying the formula for the sum of an arithmetic progression, we get that

$$
q=\frac{2}{1 \cdot 2}+\frac{2}{2 \cdot 3}+\frac{2}{3 \cdot 4}+\ldots+\frac{2}{2020 \cdot 2021}
$$

Now, using the identity $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$, we get

$$
q=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\ldots+\frac{2}{2020}-\frac{2}{2021}=\frac{2 \cdot 2020}{2021}
$$

Substituting this expression into the original fraction and simplifying it, we get the answer: 2021.