Cono Sur Olympiad

To find the sum of the sequence:

$$1 + 11 + 111 + \cdots + \underbrace{111\ldots111}_{n\text{ digits}}$$

we notice that each term in the sequence consists of digits '1' repeated a certain number of times. Specifically, the $k$-th term in the sequence is formed by $k$ digits of '1', which can be expressed as:

$$\underbrace{111\ldots1}_{k \text{ digits}} = \frac{10^k - 1}{9}$$

This is due to the fact that a number with $k$ ones can be expressed as a proper fraction in terms of powers of 10. Specifically:

$$111\ldots1 = 10^{k-1} + 10^{k-2} + \cdots + 10^1 + 1$$

This is a geometric series with first term 1 and ratio 10, so the sum is:

$$\frac{10^k - 1}{9}$$

Hence, the sum $S$ of the sequence up to $n$ terms is:

$$S = \sum_{k=1}^{n} \frac{10^k - 1}{9}$$

This can be rewritten as:

$$S = \frac{1}{9} \sum_{k=1}^{n} (10^k - 1)$$

Expanding the inner sum:

$$\sum_{k=1}^{n} (10^k - 1) = \sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1$$

The first part, $\sum_{k=1}^{n} 10^k$, is the sum of a geometric series:

$$\sum_{k=1}^{n} 10^k = 10 + 10^2 + \cdots + 10^n = 10 \frac{10^n - 1}{10 - 1} = \frac{10^{n+1} - 10}{9}$$

The second part is simply $n$, since we are summing 1 a total of $n$ times:

$$\sum_{k=1}^{n} 1 = n$$

Substituting back, we get:

$$S = \frac{1}{9} \left( \frac{10^{n+1} - 10}{9} - n \right)$$

Simplifying further, we obtain:

$$S = \frac{10^{n+1} - 10 - 9n}{81}$$

Therefore, the sum of the sequence is:

$$\boxed{\frac{10^{n+1} - 10 - 9n}{81}}$$