Task 2. (10 points) Simplify the expression $3+33+333+\ldots+\underbrace{333 . .3}_{2021}$.

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Solution. Rewrite the sum $3+33+333+\ldots+\underbrace{333 ...3}_{2021}$ as

$3 \cdot 1+3 \cdot 11+3 \cdot 111+\ldots+3 \cdot \underbrace{111 \ldots 1}_{2021}=3 \cdot \frac{10-1}{9}+3 \cdot \frac{10^{2}-1}{9}+3 \cdot \frac{10^{3}-1}{9}+\ldots+3 \cdot \frac{10^{2021}-1}{9}=$ and apply the formula for the geometric progression:

$$
\begin{aligned}
& =\frac{\left(10+10^{2}+10^{3}+\ldots+10^{2021}\right)-2021}{3}=\frac{\frac{10 \cdot\left(10^{2021}-1\right)}{9}-2021}{3} \\
& =\frac{10 \cdot\left(10^{2021}-1\right)}{27}-\frac{2021}{3}=\frac{10^{2022}-18199}{27} . \\
& \text { Answer. } \frac{10 \cdot\left(10^{2021}-1\right)}{27}-\frac{2021}{3}=\frac{10^{2022}-18199}{27} .
\end{aligned}
$$