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

$1,12,123,1234, \ldots$
$$
S_{n}=1+12+123+1234 \ldots=1+(1+11)+(1+11+111)+(1+11+111+1111)+\ldots
$$

It is known that (K. M. L. VIII, 25. 1.)

$$
1+11+111+1111+\ldots=\frac{10^{n+1}-10-9 n}{81}
$$

thus

$$
\begin{gathered}
S_{n}=\frac{10^{n+1}-10-9 n}{81}+\frac{10^{n}-10-9(n-1)}{81}+\ldots+\frac{10^{2}-10-9}{81} \\
S_{n}=\frac{1}{81}\left[10^{n+1}+10^{n}+\ldots+10^{3}+10^{2}-10 n-9\{n+(n-1)+\ldots+2+1\}\right] \\
S_{n}=\frac{1}{81}\left[10^{2} \frac{10^{n}-1}{9}-\frac{n}{2}(9 n+29)\right]
\end{gathered}
$$

(Béla Bayer, Losoncz.)

The problem was also solved by: F. Aezél, I. Bartók, G. Bogdán, A. Dessauer, A. Haar, Z. Harsányi, Gy. Hirschfeld, S. Kalmár, A. Klein, F. Kertész, G. Kertész, D. König, P. Ligeti, F. Papp, P. Pilczer, J. Pivnyik, Gv. Póka, K. Riesz, J. Sasvári, A. Schlesinger, Gy. Sümegi, H. Szmodics, J. L. Tóbiás, P. Weisz, S. Wohlstein.