Calculate $1+3+3^{2}+\ldots+3^{2020}$.
We recognize here $1^{2020}+3 \cdot 1^{2019}+3^{2} \cdot 1^{2018}+\ldots 3^{2020}$ which is a factor in the factorization of $3^{2021}-1^{2021}$. We have

$$
3^{2021}-1^{2021}=(3-1)\left(1+3+3^{2}+\ldots+3^{2020}\right)
$$

thus

$$
1+3+3^{2}+\ldots+3^{2020}=\frac{3^{2021}-1}{2}
$$