Let's determine the sum of the first $n$ terms of the following sequence:

1, 11, 111, 1111, 11111, ...
$$
\begin{gathered}
S_{n}=1+11+111+\ldots=\frac{9+99+999+\ldots}{9}= \\
=\frac{10-1+10^{2}-1+10^{3}-1+\ldots}{9}=\frac{10+10^{2}+10^{3}+\ldots+10^{n}-n}{9}= \\
=\frac{1}{9} \cdot\left[10 \frac{10^{n}-1}{9}-n\right]=\frac{10^{n+1}-10-9 n}{81} .
\end{gathered}
$$

(Bayer Béla, Losoncz.)

The problem was also solved by: Aczél F., Bartók I., Dányi I., Filkorn J., Holzmann M., Kertész F., Kertész G., König D., Lázár L., Návay L., Pilczer P., Póka Gy., Riesz M., Russo M., Scharff J., Schwarz J., Spitzer V., Sümegi Gy., Szmodics H., Tézner E., Weisz A., Wohlstein S.